A kind of asymptote for generalized alternating series
This is an interesting problem (Problem 1.6.1) from the book "Problems in Real Analysis" by Radulescu et al.:
Let $sum b_n a_n$ be a convergent series, where $a_1 geq a_2 geq ldots geq 0$, and $b_i = pm 1$ for all $i$.
Show that $(b_1 + b_2 + ldots + b_n)a_n$ converges to $0$ as $n to infty$.
I'd love to see some solutions.
real-analysis
asked Nov 29 '18 at 12:09
user97678
756
add a comment |
This is an interesting problem (Problem 1.6.1) from the book "Problems in Real Analysis" by Radulescu et al.:
Let $sum b_n a_n$ be a convergent series, where $a_1 geq a_2 geq ldots geq 0$, and $b_i = pm 1$ for all $i$.
Show that $(b_1 + b_2 + ldots + b_n)a_n$ converges to $0$ as $n to infty$.
I'd love to see some solutions.
real-analysis
asked Nov 29 '18 at 12:09
user97678
756
add a comment |
This is an interesting problem (Problem 1.6.1) from the book "Problems in Real Analysis" by Radulescu et al.:
Let $sum b_n a_n$ be a convergent series, where $a_1 geq a_2 geq ldots geq 0$, and $b_i = pm 1$ for all $i$.
Show that $(b_1 + b_2 + ldots + b_n)a_n$ converges to $0$ as $n to infty$.
I'd love to see some solutions.
real-analysis
asked Nov 29 '18 at 12:09
user97678
756
This is an interesting problem (Problem 1.6.1) from the book "Problems in Real Analysis" by Radulescu et al.:
Let $sum b_n a_n$ be a convergent series, where $a_1 geq a_2 geq ldots geq 0$, and $b_i = pm 1$ for all $i$.
Show that $(b_1 + b_2 + ldots + b_n)a_n$ converges to $0$ as $n to infty$.
I'd love to see some solutions.
real-analysis
real-analysis
asked Nov 29 '18 at 12:09
user97678
756
asked Nov 29 '18 at 12:09
user97678
756
asked Nov 29 '18 at 12:09
user97678
756
asked Nov 29 '18 at 12:09
user97678
756
asked Nov 29 '18 at 12:09
user97678
756
756
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
There is a simple proof, without to many technical steps, with the help of Kronecker's lemma. In fact, it is a special case of Kronecker's lemma.
If $alpha_k in mathbb{R}$ and $(beta_k)_k$ is an increasing sequence of positive real numbers with $beta_k rightarrow infty$ such that $sum_{k=1}^infty alpha_k$ is convergent, then $frac{1}{beta_n} sum_{k=1}^n beta_k alpha_k rightarrow 0$ for $n rightarrow infty$.
Proof: Write $S_n := sum_{k=1}^n alpha_k$, then we have
$$frac{1}{beta_n} sum_{k=1}^n beta_k alpha_k = frac{1}{beta_n} sum_{k=1}^n beta_k (S_k -S_{k-1}) = S_n - frac{1}{beta_n} sum_{k=1}^{n-1} (beta_{k+1}-beta_k) S_k $$
Since $S_n rightarrow s$, it remains to prove that $ frac{1}{beta_n} sum_{k=1}^{n-1} (beta_{k+1}-beta_k) S_k rightarrow s$. Take $|S_n -s| < varepsilon$ for all $n ge N$. Then we have for all $n > N$ that
begin{align}
&frac{1}{beta_n} sum_{k=1}^{n-1} (beta_{k+1}-beta_k) S_k-(1- beta_1/beta_n)s \
=& frac{1}{beta_n} sum_{k=1}^{n-1} (beta_{k+1}-beta_k) (S_k-s) \
=&frac{1}{beta_n} sum_{k=1}^{N-1} (beta_{k+1}-beta_k) (S_k-s) + frac{1}{beta_n} sum_{k=N}^{n-1} (beta_{k+1}-beta_k) (S_k-s)
end{align}
We have $(1- beta_1/beta_n)s rightarrow s$ and $frac{1}{beta_n} sum_{k=1}^{N-1} (beta_{k+1}-beta_k) (S_k-s) rightarrow 0$ for fixed $N$. On the other hand
$$left|frac{1}{beta_n} sum_{k=N}^{n} (beta_{k+1}-beta_k) (S_k-s) right| le varepsilon frac{1}{beta_n} (beta_n - beta_N) le varepsilon M $$
This proves Kronecker's lemma.
Now, we would like to apply the lemma as follows. Our first observation is that $a_n rightarrow 0$ for $n rightarrow infty$, because $|a_n b_b| = |a_n|$ and $b_n a_n rightarrow 0$ for $n rightarrow infty$, since the sum converges. Thus, we can take $beta_n = a_k^{-1}$ and this sequence satisfies Kronecker's lemma, because $(a_n)_n$ is also decreasing. Since $sum_{k=1}^infty (b_k a_k)$ is convergent (we take $alpha_k = b_k a_k$), we get that
$$a_m sum_{k=1}^m b_k = frac{1}{beta_m} sum_{k=1}^m alpha_k beta_k rightarrow 0.$$
answered Nov 29 '18 at 12:57
p4sch
4,770217
Why does a $k$ fulfilling your inequality (2) exist?
– user97678
Nov 29 '18 at 13:40
We have $n_k rightarrow infty$ for $k rightarrow infty$. On the other hand, we can use that the series convegences, i.e. is a Cauchy-sequence. Thus, for any $varepsilon >0$ there exists $N in mathbb{N}$ with $|sum_{i=n}^m b_i a_i| < varepsilon$ for all $m >n ge N$. Specially, we can take (for $k$ large enough) $n = n_k$ and $m = (n_{k+1}-1) wedge m$.
– p4sch
Nov 29 '18 at 13:44
I can see why due to Cauchy it has to be $< epsilon$, but why is the left part of the inequality correct? I'm not sure why $n_{k+1} wedge (m-n_k)a_m$ has to be smaller than the sum
– user97678
Nov 29 '18 at 13:48
Note that $(a_n)$ is decreasing and that we have $b_i =1$ or $b_i = -1$ for all terms in the sum. Thus the sum is just $left| sum_{i=n_l}^{(n_{l+1}-1) wedge m} a_i right| ge a_m [(n_{l+1}-1) wedge m-n_l+1]$.
– p4sch
Nov 29 '18 at 13:59
But why is $b_i$'s sign fixed for all terms in the sum? Isn't your $m$ unbounded? You can pick it high enough that the $b_i$s start changing signs, no?
– user97678
Nov 29 '18 at 14:01
|
show 2 more comments
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3018535%2fa-kind-of-asymptote-for-generalized-alternating-series%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
There is a simple proof, without to many technical steps, with the help of Kronecker's lemma. In fact, it is a special case of Kronecker's lemma.
If $alpha_k in mathbb{R}$ and $(beta_k)_k$ is an increasing sequence of positive real numbers with $beta_k rightarrow infty$ such that $sum_{k=1}^infty alpha_k$ is convergent, then $frac{1}{beta_n} sum_{k=1}^n beta_k alpha_k rightarrow 0$ for $n rightarrow infty$.
Proof: Write $S_n := sum_{k=1}^n alpha_k$, then we have
$$frac{1}{beta_n} sum_{k=1}^n beta_k alpha_k = frac{1}{beta_n} sum_{k=1}^n beta_k (S_k -S_{k-1}) = S_n - frac{1}{beta_n} sum_{k=1}^{n-1} (beta_{k+1}-beta_k) S_k $$
Since $S_n rightarrow s$, it remains to prove that $ frac{1}{beta_n} sum_{k=1}^{n-1} (beta_{k+1}-beta_k) S_k rightarrow s$. Take $|S_n -s| < varepsilon$ for all $n ge N$. Then we have for all $n > N$ that
begin{align}
&frac{1}{beta_n} sum_{k=1}^{n-1} (beta_{k+1}-beta_k) S_k-(1- beta_1/beta_n)s \
=& frac{1}{beta_n} sum_{k=1}^{n-1} (beta_{k+1}-beta_k) (S_k-s) \
=&frac{1}{beta_n} sum_{k=1}^{N-1} (beta_{k+1}-beta_k) (S_k-s) + frac{1}{beta_n} sum_{k=N}^{n-1} (beta_{k+1}-beta_k) (S_k-s)
end{align}
We have $(1- beta_1/beta_n)s rightarrow s$ and $frac{1}{beta_n} sum_{k=1}^{N-1} (beta_{k+1}-beta_k) (S_k-s) rightarrow 0$ for fixed $N$. On the other hand
$$left|frac{1}{beta_n} sum_{k=N}^{n} (beta_{k+1}-beta_k) (S_k-s) right| le varepsilon frac{1}{beta_n} (beta_n - beta_N) le varepsilon M $$
This proves Kronecker's lemma.
Now, we would like to apply the lemma as follows. Our first observation is that $a_n rightarrow 0$ for $n rightarrow infty$, because $|a_n b_b| = |a_n|$ and $b_n a_n rightarrow 0$ for $n rightarrow infty$, since the sum converges. Thus, we can take $beta_n = a_k^{-1}$ and this sequence satisfies Kronecker's lemma, because $(a_n)_n$ is also decreasing. Since $sum_{k=1}^infty (b_k a_k)$ is convergent (we take $alpha_k = b_k a_k$), we get that
$$a_m sum_{k=1}^m b_k = frac{1}{beta_m} sum_{k=1}^m alpha_k beta_k rightarrow 0.$$
answered Nov 29 '18 at 12:57
p4sch
4,770217
Why does a $k$ fulfilling your inequality (2) exist?
– user97678
Nov 29 '18 at 13:40
We have $n_k rightarrow infty$ for $k rightarrow infty$. On the other hand, we can use that the series convegences, i.e. is a Cauchy-sequence. Thus, for any $varepsilon >0$ there exists $N in mathbb{N}$ with $|sum_{i=n}^m b_i a_i| < varepsilon$ for all $m >n ge N$. Specially, we can take (for $k$ large enough) $n = n_k$ and $m = (n_{k+1}-1) wedge m$.
– p4sch
Nov 29 '18 at 13:44
I can see why due to Cauchy it has to be $< epsilon$, but why is the left part of the inequality correct? I'm not sure why $n_{k+1} wedge (m-n_k)a_m$ has to be smaller than the sum
– user97678
Nov 29 '18 at 13:48
Note that $(a_n)$ is decreasing and that we have $b_i =1$ or $b_i = -1$ for all terms in the sum. Thus the sum is just $left| sum_{i=n_l}^{(n_{l+1}-1) wedge m} a_i right| ge a_m [(n_{l+1}-1) wedge m-n_l+1]$.
– p4sch
Nov 29 '18 at 13:59
But why is $b_i$'s sign fixed for all terms in the sum? Isn't your $m$ unbounded? You can pick it high enough that the $b_i$s start changing signs, no?
– user97678
Nov 29 '18 at 14:01
|
show 2 more comments
There is a simple proof, without to many technical steps, with the help of Kronecker's lemma. In fact, it is a special case of Kronecker's lemma.
If $alpha_k in mathbb{R}$ and $(beta_k)_k$ is an increasing sequence of positive real numbers with $beta_k rightarrow infty$ such that $sum_{k=1}^infty alpha_k$ is convergent, then $frac{1}{beta_n} sum_{k=1}^n beta_k alpha_k rightarrow 0$ for $n rightarrow infty$.
Proof: Write $S_n := sum_{k=1}^n alpha_k$, then we have
$$frac{1}{beta_n} sum_{k=1}^n beta_k alpha_k = frac{1}{beta_n} sum_{k=1}^n beta_k (S_k -S_{k-1}) = S_n - frac{1}{beta_n} sum_{k=1}^{n-1} (beta_{k+1}-beta_k) S_k $$
Since $S_n rightarrow s$, it remains to prove that $ frac{1}{beta_n} sum_{k=1}^{n-1} (beta_{k+1}-beta_k) S_k rightarrow s$. Take $|S_n -s| < varepsilon$ for all $n ge N$. Then we have for all $n > N$ that
begin{align}
&frac{1}{beta_n} sum_{k=1}^{n-1} (beta_{k+1}-beta_k) S_k-(1- beta_1/beta_n)s \
=& frac{1}{beta_n} sum_{k=1}^{n-1} (beta_{k+1}-beta_k) (S_k-s) \
=&frac{1}{beta_n} sum_{k=1}^{N-1} (beta_{k+1}-beta_k) (S_k-s) + frac{1}{beta_n} sum_{k=N}^{n-1} (beta_{k+1}-beta_k) (S_k-s)
end{align}
We have $(1- beta_1/beta_n)s rightarrow s$ and $frac{1}{beta_n} sum_{k=1}^{N-1} (beta_{k+1}-beta_k) (S_k-s) rightarrow 0$ for fixed $N$. On the other hand
$$left|frac{1}{beta_n} sum_{k=N}^{n} (beta_{k+1}-beta_k) (S_k-s) right| le varepsilon frac{1}{beta_n} (beta_n - beta_N) le varepsilon M $$
This proves Kronecker's lemma.
Now, we would like to apply the lemma as follows. Our first observation is that $a_n rightarrow 0$ for $n rightarrow infty$, because $|a_n b_b| = |a_n|$ and $b_n a_n rightarrow 0$ for $n rightarrow infty$, since the sum converges. Thus, we can take $beta_n = a_k^{-1}$ and this sequence satisfies Kronecker's lemma, because $(a_n)_n$ is also decreasing. Since $sum_{k=1}^infty (b_k a_k)$ is convergent (we take $alpha_k = b_k a_k$), we get that
$$a_m sum_{k=1}^m b_k = frac{1}{beta_m} sum_{k=1}^m alpha_k beta_k rightarrow 0.$$
answered Nov 29 '18 at 12:57
p4sch
4,770217
Why does a $k$ fulfilling your inequality (2) exist?
– user97678
Nov 29 '18 at 13:40
We have $n_k rightarrow infty$ for $k rightarrow infty$. On the other hand, we can use that the series convegences, i.e. is a Cauchy-sequence. Thus, for any $varepsilon >0$ there exists $N in mathbb{N}$ with $|sum_{i=n}^m b_i a_i| < varepsilon$ for all $m >n ge N$. Specially, we can take (for $k$ large enough) $n = n_k$ and $m = (n_{k+1}-1) wedge m$.
– p4sch
Nov 29 '18 at 13:44
I can see why due to Cauchy it has to be $< epsilon$, but why is the left part of the inequality correct? I'm not sure why $n_{k+1} wedge (m-n_k)a_m$ has to be smaller than the sum
– user97678
Nov 29 '18 at 13:48
Note that $(a_n)$ is decreasing and that we have $b_i =1$ or $b_i = -1$ for all terms in the sum. Thus the sum is just $left| sum_{i=n_l}^{(n_{l+1}-1) wedge m} a_i right| ge a_m [(n_{l+1}-1) wedge m-n_l+1]$.
– p4sch
Nov 29 '18 at 13:59
But why is $b_i$'s sign fixed for all terms in the sum? Isn't your $m$ unbounded? You can pick it high enough that the $b_i$s start changing signs, no?
– user97678
Nov 29 '18 at 14:01
|
show 2 more comments
There is a simple proof, without to many technical steps, with the help of Kronecker's lemma. In fact, it is a special case of Kronecker's lemma.
If $alpha_k in mathbb{R}$ and $(beta_k)_k$ is an increasing sequence of positive real numbers with $beta_k rightarrow infty$ such that $sum_{k=1}^infty alpha_k$ is convergent, then $frac{1}{beta_n} sum_{k=1}^n beta_k alpha_k rightarrow 0$ for $n rightarrow infty$.
Proof: Write $S_n := sum_{k=1}^n alpha_k$, then we have
$$frac{1}{beta_n} sum_{k=1}^n beta_k alpha_k = frac{1}{beta_n} sum_{k=1}^n beta_k (S_k -S_{k-1}) = S_n - frac{1}{beta_n} sum_{k=1}^{n-1} (beta_{k+1}-beta_k) S_k $$
Since $S_n rightarrow s$, it remains to prove that $ frac{1}{beta_n} sum_{k=1}^{n-1} (beta_{k+1}-beta_k) S_k rightarrow s$. Take $|S_n -s| < varepsilon$ for all $n ge N$. Then we have for all $n > N$ that
begin{align}
&frac{1}{beta_n} sum_{k=1}^{n-1} (beta_{k+1}-beta_k) S_k-(1- beta_1/beta_n)s \
=& frac{1}{beta_n} sum_{k=1}^{n-1} (beta_{k+1}-beta_k) (S_k-s) \
=&frac{1}{beta_n} sum_{k=1}^{N-1} (beta_{k+1}-beta_k) (S_k-s) + frac{1}{beta_n} sum_{k=N}^{n-1} (beta_{k+1}-beta_k) (S_k-s)
end{align}
We have $(1- beta_1/beta_n)s rightarrow s$ and $frac{1}{beta_n} sum_{k=1}^{N-1} (beta_{k+1}-beta_k) (S_k-s) rightarrow 0$ for fixed $N$. On the other hand
$$left|frac{1}{beta_n} sum_{k=N}^{n} (beta_{k+1}-beta_k) (S_k-s) right| le varepsilon frac{1}{beta_n} (beta_n - beta_N) le varepsilon M $$
This proves Kronecker's lemma.
Now, we would like to apply the lemma as follows. Our first observation is that $a_n rightarrow 0$ for $n rightarrow infty$, because $|a_n b_b| = |a_n|$ and $b_n a_n rightarrow 0$ for $n rightarrow infty$, since the sum converges. Thus, we can take $beta_n = a_k^{-1}$ and this sequence satisfies Kronecker's lemma, because $(a_n)_n$ is also decreasing. Since $sum_{k=1}^infty (b_k a_k)$ is convergent (we take $alpha_k = b_k a_k$), we get that
$$a_m sum_{k=1}^m b_k = frac{1}{beta_m} sum_{k=1}^m alpha_k beta_k rightarrow 0.$$
answered Nov 29 '18 at 12:57
p4sch
4,770217
There is a simple proof, without to many technical steps, with the help of Kronecker's lemma. In fact, it is a special case of Kronecker's lemma.
If $alpha_k in mathbb{R}$ and $(beta_k)_k$ is an increasing sequence of positive real numbers with $beta_k rightarrow infty$ such that $sum_{k=1}^infty alpha_k$ is convergent, then $frac{1}{beta_n} sum_{k=1}^n beta_k alpha_k rightarrow 0$ for $n rightarrow infty$.
Proof: Write $S_n := sum_{k=1}^n alpha_k$, then we have
$$frac{1}{beta_n} sum_{k=1}^n beta_k alpha_k = frac{1}{beta_n} sum_{k=1}^n beta_k (S_k -S_{k-1}) = S_n - frac{1}{beta_n} sum_{k=1}^{n-1} (beta_{k+1}-beta_k) S_k $$
Since $S_n rightarrow s$, it remains to prove that $ frac{1}{beta_n} sum_{k=1}^{n-1} (beta_{k+1}-beta_k) S_k rightarrow s$. Take $|S_n -s| < varepsilon$ for all $n ge N$. Then we have for all $n > N$ that
begin{align}
&frac{1}{beta_n} sum_{k=1}^{n-1} (beta_{k+1}-beta_k) S_k-(1- beta_1/beta_n)s \
=& frac{1}{beta_n} sum_{k=1}^{n-1} (beta_{k+1}-beta_k) (S_k-s) \
=&frac{1}{beta_n} sum_{k=1}^{N-1} (beta_{k+1}-beta_k) (S_k-s) + frac{1}{beta_n} sum_{k=N}^{n-1} (beta_{k+1}-beta_k) (S_k-s)
end{align}
We have $(1- beta_1/beta_n)s rightarrow s$ and $frac{1}{beta_n} sum_{k=1}^{N-1} (beta_{k+1}-beta_k) (S_k-s) rightarrow 0$ for fixed $N$. On the other hand
$$left|frac{1}{beta_n} sum_{k=N}^{n} (beta_{k+1}-beta_k) (S_k-s) right| le varepsilon frac{1}{beta_n} (beta_n - beta_N) le varepsilon M $$
This proves Kronecker's lemma.
Now, we would like to apply the lemma as follows. Our first observation is that $a_n rightarrow 0$ for $n rightarrow infty$, because $|a_n b_b| = |a_n|$ and $b_n a_n rightarrow 0$ for $n rightarrow infty$, since the sum converges. Thus, we can take $beta_n = a_k^{-1}$ and this sequence satisfies Kronecker's lemma, because $(a_n)_n$ is also decreasing. Since $sum_{k=1}^infty (b_k a_k)$ is convergent (we take $alpha_k = b_k a_k$), we get that
$$a_m sum_{k=1}^m b_k = frac{1}{beta_m} sum_{k=1}^m alpha_k beta_k rightarrow 0.$$
answered Nov 29 '18 at 12:57
p4sch
4,770217
edited Nov 29 '18 at 15:59
answered Nov 29 '18 at 12:57
p4sch
4,770217
answered Nov 29 '18 at 12:57
p4sch
4,770217
answered Nov 29 '18 at 12:57
p4sch
4,770217
4,770217
Why does a $k$ fulfilling your inequality (2) exist?
– user97678
Nov 29 '18 at 13:40
We have $n_k rightarrow infty$ for $k rightarrow infty$. On the other hand, we can use that the series convegences, i.e. is a Cauchy-sequence. Thus, for any $varepsilon >0$ there exists $N in mathbb{N}$ with $|sum_{i=n}^m b_i a_i| < varepsilon$ for all $m >n ge N$. Specially, we can take (for $k$ large enough) $n = n_k$ and $m = (n_{k+1}-1) wedge m$.
– p4sch
Nov 29 '18 at 13:44
I can see why due to Cauchy it has to be $< epsilon$, but why is the left part of the inequality correct? I'm not sure why $n_{k+1} wedge (m-n_k)a_m$ has to be smaller than the sum
– user97678
Nov 29 '18 at 13:48
Note that $(a_n)$ is decreasing and that we have $b_i =1$ or $b_i = -1$ for all terms in the sum. Thus the sum is just $left| sum_{i=n_l}^{(n_{l+1}-1) wedge m} a_i right| ge a_m [(n_{l+1}-1) wedge m-n_l+1]$.
– p4sch
Nov 29 '18 at 13:59
But why is $b_i$'s sign fixed for all terms in the sum? Isn't your $m$ unbounded? You can pick it high enough that the $b_i$s start changing signs, no?
– user97678
Nov 29 '18 at 14:01
|
show 2 more comments
Why does a $k$ fulfilling your inequality (2) exist?
– user97678
Nov 29 '18 at 13:40
We have $n_k rightarrow infty$ for $k rightarrow infty$. On the other hand, we can use that the series convegences, i.e. is a Cauchy-sequence. Thus, for any $varepsilon >0$ there exists $N in mathbb{N}$ with $|sum_{i=n}^m b_i a_i| < varepsilon$ for all $m >n ge N$. Specially, we can take (for $k$ large enough) $n = n_k$ and $m = (n_{k+1}-1) wedge m$.
– p4sch
Nov 29 '18 at 13:44
I can see why due to Cauchy it has to be $< epsilon$, but why is the left part of the inequality correct? I'm not sure why $n_{k+1} wedge (m-n_k)a_m$ has to be smaller than the sum
– user97678
Nov 29 '18 at 13:48
Note that $(a_n)$ is decreasing and that we have $b_i =1$ or $b_i = -1$ for all terms in the sum. Thus the sum is just $left| sum_{i=n_l}^{(n_{l+1}-1) wedge m} a_i right| ge a_m [(n_{l+1}-1) wedge m-n_l+1]$.
– p4sch
Nov 29 '18 at 13:59
But why is $b_i$'s sign fixed for all terms in the sum? Isn't your $m$ unbounded? You can pick it high enough that the $b_i$s start changing signs, no?
– user97678
Nov 29 '18 at 14:01
Why does a $k$ fulfilling your inequality (2) exist?
– user97678
Nov 29 '18 at 13:40
Why does a $k$ fulfilling your inequality (2) exist?
– user97678
Nov 29 '18 at 13:40
We have $n_k rightarrow infty$ for $k rightarrow infty$. On the other hand, we can use that the series convegences, i.e. is a Cauchy-sequence. Thus, for any $varepsilon >0$ there exists $N in mathbb{N}$ with $|sum_{i=n}^m b_i a_i| < varepsilon$ for all $m >n ge N$. Specially, we can take (for $k$ large enough) $n = n_k$ and $m = (n_{k+1}-1) wedge m$.
– p4sch
Nov 29 '18 at 13:44
We have $n_k rightarrow infty$ for $k rightarrow infty$. On the other hand, we can use that the series convegences, i.e. is a Cauchy-sequence. Thus, for any $varepsilon >0$ there exists $N in mathbb{N}$ with $|sum_{i=n}^m b_i a_i| < varepsilon$ for all $m >n ge N$. Specially, we can take (for $k$ large enough) $n = n_k$ and $m = (n_{k+1}-1) wedge m$.
– p4sch
Nov 29 '18 at 13:44
I can see why due to Cauchy it has to be $< epsilon$, but why is the left part of the inequality correct? I'm not sure why $n_{k+1} wedge (m-n_k)a_m$ has to be smaller than the sum
– user97678
Nov 29 '18 at 13:48
I can see why due to Cauchy it has to be $< epsilon$, but why is the left part of the inequality correct? I'm not sure why $n_{k+1} wedge (m-n_k)a_m$ has to be smaller than the sum
– user97678
Nov 29 '18 at 13:48
Note that $(a_n)$ is decreasing and that we have $b_i =1$ or $b_i = -1$ for all terms in the sum. Thus the sum is just $left| sum_{i=n_l}^{(n_{l+1}-1) wedge m} a_i right| ge a_m [(n_{l+1}-1) wedge m-n_l+1]$.
– p4sch
Nov 29 '18 at 13:59
Note that $(a_n)$ is decreasing and that we have $b_i =1$ or $b_i = -1$ for all terms in the sum. Thus the sum is just $left| sum_{i=n_l}^{(n_{l+1}-1) wedge m} a_i right| ge a_m [(n_{l+1}-1) wedge m-n_l+1]$.
– p4sch
Nov 29 '18 at 13:59
But why is $b_i$'s sign fixed for all terms in the sum? Isn't your $m$ unbounded? You can pick it high enough that the $b_i$s start changing signs, no?
– user97678
Nov 29 '18 at 14:01
But why is $b_i$'s sign fixed for all terms in the sum? Isn't your $m$ unbounded? You can pick it high enough that the $b_i$s start changing signs, no?
– user97678
Nov 29 '18 at 14:01
|
show 2 more comments
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3018535%2fa-kind-of-asymptote-for-generalized-alternating-series%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
