Analyzing $sum^infty _{n=1}{frac{1}{p^n - q^n}}, 0lt qlt p$ [closed]
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I am told to analyze the convergence of $sum^infty _{n=1}{frac{1}{p^n - q^n}}, 0lt qlt p$ depending on the values of $p$ and $q$. The only thing I have concluded is that ovbiously, the series is a positive term series. I do not know which criteria to use in this case. Thanks in advance.
real-analysis calculus sequences-and-series
edited Jan 3 at 17:52
Foobaz John
22.9k41552
asked Jan 3 at 17:30
AndarrkorAndarrkor
466
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closed as off-topic by Did, Strants, rtybase, Namaste, José Carlos Santos Jan 3 at 23:58
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, Strants, rtybase, Namaste, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I am told to analyze the convergence of $sum^infty _{n=1}{frac{1}{p^n - q^n}}, 0lt qlt p$ depending on the values of $p$ and $q$. The only thing I have concluded is that ovbiously, the series is a positive term series. I do not know which criteria to use in this case. Thanks in advance.
real-analysis calculus sequences-and-series
edited Jan 3 at 17:52
Foobaz John
22.9k41552
asked Jan 3 at 17:30
AndarrkorAndarrkor
466
$endgroup$
closed as off-topic by Did, Strants, rtybase, Namaste, José Carlos Santos Jan 3 at 23:58
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, Strants, rtybase, Namaste, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Faced with such a problem, I would experiment. Try different values of p, q, and n (e.g. p=3 or 5, q = 2, n =5. Look for patterns in the behavior of the sums of the fractions. Form hypotheses around the observed behavior. Prove the hypotheses. Form hypotheses around convergence. Try to prove these hypostheses. Then, if you still can't complete the problem, edit your post to show all of your work.
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– user2661923
Jan 3 at 17:39
add a comment |
$begingroup$
I am told to analyze the convergence of $sum^infty _{n=1}{frac{1}{p^n - q^n}}, 0lt qlt p$ depending on the values of $p$ and $q$. The only thing I have concluded is that ovbiously, the series is a positive term series. I do not know which criteria to use in this case. Thanks in advance.
real-analysis calculus sequences-and-series
edited Jan 3 at 17:52
Foobaz John
22.9k41552
asked Jan 3 at 17:30
AndarrkorAndarrkor
466
$endgroup$
I am told to analyze the convergence of $sum^infty _{n=1}{frac{1}{p^n - q^n}}, 0lt qlt p$ depending on the values of $p$ and $q$. The only thing I have concluded is that ovbiously, the series is a positive term series. I do not know which criteria to use in this case. Thanks in advance.
real-analysis calculus sequences-and-series
real-analysis calculus sequences-and-series
edited Jan 3 at 17:52
Foobaz John
22.9k41552
asked Jan 3 at 17:30
AndarrkorAndarrkor
466
edited Jan 3 at 17:52
Foobaz John
22.9k41552
asked Jan 3 at 17:30
AndarrkorAndarrkor
466
edited Jan 3 at 17:52
Foobaz John
22.9k41552
edited Jan 3 at 17:52
Foobaz John
22.9k41552
edited Jan 3 at 17:52
Foobaz John
22.9k41552
22.9k41552
asked Jan 3 at 17:30
AndarrkorAndarrkor
466
asked Jan 3 at 17:30
AndarrkorAndarrkor
466
asked Jan 3 at 17:30
AndarrkorAndarrkor
466
466
closed as off-topic by Did, Strants, rtybase, Namaste, José Carlos Santos Jan 3 at 23:58
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, Strants, rtybase, Namaste, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Did, Strants, rtybase, Namaste, José Carlos Santos Jan 3 at 23:58
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, Strants, rtybase, Namaste, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Faced with such a problem, I would experiment. Try different values of p, q, and n (e.g. p=3 or 5, q = 2, n =5. Look for patterns in the behavior of the sums of the fractions. Form hypotheses around the observed behavior. Prove the hypotheses. Form hypotheses around convergence. Try to prove these hypostheses. Then, if you still can't complete the problem, edit your post to show all of your work.
$endgroup$
– user2661923
Jan 3 at 17:39
add a comment |
$begingroup$
Faced with such a problem, I would experiment. Try different values of p, q, and n (e.g. p=3 or 5, q = 2, n =5. Look for patterns in the behavior of the sums of the fractions. Form hypotheses around the observed behavior. Prove the hypotheses. Form hypotheses around convergence. Try to prove these hypostheses. Then, if you still can't complete the problem, edit your post to show all of your work.
$endgroup$
– user2661923
Jan 3 at 17:39
$begingroup$
Faced with such a problem, I would experiment. Try different values of p, q, and n (e.g. p=3 or 5, q = 2, n =5. Look for patterns in the behavior of the sums of the fractions. Form hypotheses around the observed behavior. Prove the hypotheses. Form hypotheses around convergence. Try to prove these hypostheses. Then, if you still can't complete the problem, edit your post to show all of your work.
$endgroup$
– user2661923
Jan 3 at 17:39
$begingroup$
Faced with such a problem, I would experiment. Try different values of p, q, and n (e.g. p=3 or 5, q = 2, n =5. Look for patterns in the behavior of the sums of the fractions. Form hypotheses around the observed behavior. Prove the hypotheses. Form hypotheses around convergence. Try to prove these hypostheses. Then, if you still can't complete the problem, edit your post to show all of your work.
$endgroup$
– user2661923
Jan 3 at 17:39
add a comment |
2 Answers
2
active
oldest
votes
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Observe that
$$sum_{n=1}^inftyfrac{1}{p^n-q^n}=frac{1}{p-q}sum_{n=1}^inftyfrac{1}{p^{n-1}+p^{n-2}q+cdots+pq^{n-2}+q^{n-1}}$$
We can apply the comparison test to $dfrac{1}{p^{n-1}}$ and $dfrac{1}{q^{n-1}}$ to see that the series converges whenever $p>1$ or $q>1$.
If both $pleq 1$ and $qleq 1$, then it is not the case that $dfrac 1{(p^n-q^n)}to 0$, hence the series diverges.
edited Jan 3 at 23:35
Namaste
1
answered Jan 3 at 17:46
Ben WBen W
2,722818
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add a comment |
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Write the general term $a_n$ of the series as
$$
0lefrac{m}{p^n}leq a_n=frac{1}{p^n(1-(q/p)^n)}leq frac{M}{p^n}
$$
since $frac{1}{1-(q/p)^n}to 1$ for some $m, M>0$. The above inequality should allow you to deduce convergence depending on whether $p>1$ or $p<1$.
For the case with $p=1$, the general term of the series is
$$
a_n=frac{1}{1-q^n}
$$
What can we say about $lim_{nto infty} a_n$ if $q<1$.
answered Jan 3 at 17:47
Foobaz JohnFoobaz John
22.9k41552
$endgroup$
$begingroup$
How did you delimit the series? Thanks for answering.
$endgroup$
– Andarrkor
Jan 3 at 18:04
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Observe that
$$sum_{n=1}^inftyfrac{1}{p^n-q^n}=frac{1}{p-q}sum_{n=1}^inftyfrac{1}{p^{n-1}+p^{n-2}q+cdots+pq^{n-2}+q^{n-1}}$$
We can apply the comparison test to $dfrac{1}{p^{n-1}}$ and $dfrac{1}{q^{n-1}}$ to see that the series converges whenever $p>1$ or $q>1$.
If both $pleq 1$ and $qleq 1$, then it is not the case that $dfrac 1{(p^n-q^n)}to 0$, hence the series diverges.
edited Jan 3 at 23:35
Namaste
1
answered Jan 3 at 17:46
Ben WBen W
2,722818
$endgroup$
add a comment |
$begingroup$
Observe that
$$sum_{n=1}^inftyfrac{1}{p^n-q^n}=frac{1}{p-q}sum_{n=1}^inftyfrac{1}{p^{n-1}+p^{n-2}q+cdots+pq^{n-2}+q^{n-1}}$$
We can apply the comparison test to $dfrac{1}{p^{n-1}}$ and $dfrac{1}{q^{n-1}}$ to see that the series converges whenever $p>1$ or $q>1$.
If both $pleq 1$ and $qleq 1$, then it is not the case that $dfrac 1{(p^n-q^n)}to 0$, hence the series diverges.
edited Jan 3 at 23:35
Namaste
1
answered Jan 3 at 17:46
Ben WBen W
2,722818
$endgroup$
add a comment |
$begingroup$
Observe that
$$sum_{n=1}^inftyfrac{1}{p^n-q^n}=frac{1}{p-q}sum_{n=1}^inftyfrac{1}{p^{n-1}+p^{n-2}q+cdots+pq^{n-2}+q^{n-1}}$$
We can apply the comparison test to $dfrac{1}{p^{n-1}}$ and $dfrac{1}{q^{n-1}}$ to see that the series converges whenever $p>1$ or $q>1$.
If both $pleq 1$ and $qleq 1$, then it is not the case that $dfrac 1{(p^n-q^n)}to 0$, hence the series diverges.
edited Jan 3 at 23:35
Namaste
1
answered Jan 3 at 17:46
Ben WBen W
2,722818
$endgroup$
Observe that
$$sum_{n=1}^inftyfrac{1}{p^n-q^n}=frac{1}{p-q}sum_{n=1}^inftyfrac{1}{p^{n-1}+p^{n-2}q+cdots+pq^{n-2}+q^{n-1}}$$
We can apply the comparison test to $dfrac{1}{p^{n-1}}$ and $dfrac{1}{q^{n-1}}$ to see that the series converges whenever $p>1$ or $q>1$.
If both $pleq 1$ and $qleq 1$, then it is not the case that $dfrac 1{(p^n-q^n)}to 0$, hence the series diverges.
edited Jan 3 at 23:35
Namaste
1
answered Jan 3 at 17:46
Ben WBen W
2,722818
edited Jan 3 at 23:35
Namaste
1
edited Jan 3 at 23:35
Namaste
1
edited Jan 3 at 23:35
Namaste
1
1
answered Jan 3 at 17:46
Ben WBen W
2,722818
answered Jan 3 at 17:46
Ben WBen W
2,722818
answered Jan 3 at 17:46
Ben WBen W
2,722818
2,722818
add a comment |
add a comment |
$begingroup$
Write the general term $a_n$ of the series as
$$
0lefrac{m}{p^n}leq a_n=frac{1}{p^n(1-(q/p)^n)}leq frac{M}{p^n}
$$
since $frac{1}{1-(q/p)^n}to 1$ for some $m, M>0$. The above inequality should allow you to deduce convergence depending on whether $p>1$ or $p<1$.
For the case with $p=1$, the general term of the series is
$$
a_n=frac{1}{1-q^n}
$$
What can we say about $lim_{nto infty} a_n$ if $q<1$.
answered Jan 3 at 17:47
Foobaz JohnFoobaz John
22.9k41552
$endgroup$
$begingroup$
How did you delimit the series? Thanks for answering.
$endgroup$
– Andarrkor
Jan 3 at 18:04
add a comment |
$begingroup$
Write the general term $a_n$ of the series as
$$
0lefrac{m}{p^n}leq a_n=frac{1}{p^n(1-(q/p)^n)}leq frac{M}{p^n}
$$
since $frac{1}{1-(q/p)^n}to 1$ for some $m, M>0$. The above inequality should allow you to deduce convergence depending on whether $p>1$ or $p<1$.
For the case with $p=1$, the general term of the series is
$$
a_n=frac{1}{1-q^n}
$$
What can we say about $lim_{nto infty} a_n$ if $q<1$.
answered Jan 3 at 17:47
Foobaz JohnFoobaz John
22.9k41552
$endgroup$
$begingroup$
How did you delimit the series? Thanks for answering.
$endgroup$
– Andarrkor
Jan 3 at 18:04
add a comment |
$begingroup$
Write the general term $a_n$ of the series as
$$
0lefrac{m}{p^n}leq a_n=frac{1}{p^n(1-(q/p)^n)}leq frac{M}{p^n}
$$
since $frac{1}{1-(q/p)^n}to 1$ for some $m, M>0$. The above inequality should allow you to deduce convergence depending on whether $p>1$ or $p<1$.
For the case with $p=1$, the general term of the series is
$$
a_n=frac{1}{1-q^n}
$$
What can we say about $lim_{nto infty} a_n$ if $q<1$.
answered Jan 3 at 17:47
Foobaz JohnFoobaz John
22.9k41552
$endgroup$
Write the general term $a_n$ of the series as
$$
0lefrac{m}{p^n}leq a_n=frac{1}{p^n(1-(q/p)^n)}leq frac{M}{p^n}
$$
since $frac{1}{1-(q/p)^n}to 1$ for some $m, M>0$. The above inequality should allow you to deduce convergence depending on whether $p>1$ or $p<1$.
For the case with $p=1$, the general term of the series is
$$
a_n=frac{1}{1-q^n}
$$
What can we say about $lim_{nto infty} a_n$ if $q<1$.
answered Jan 3 at 17:47
Foobaz JohnFoobaz John
22.9k41552
answered Jan 3 at 17:47
Foobaz JohnFoobaz John
22.9k41552
answered Jan 3 at 17:47
Foobaz JohnFoobaz John
22.9k41552
answered Jan 3 at 17:47
Foobaz JohnFoobaz John
22.9k41552
22.9k41552
$begingroup$
How did you delimit the series? Thanks for answering.
$endgroup$
– Andarrkor
Jan 3 at 18:04
add a comment |
$begingroup$
How did you delimit the series? Thanks for answering.
$endgroup$
– Andarrkor
Jan 3 at 18:04
$begingroup$
How did you delimit the series? Thanks for answering.
$endgroup$
– Andarrkor
Jan 3 at 18:04
$begingroup$
How did you delimit the series? Thanks for answering.
$endgroup$
– Andarrkor
Jan 3 at 18:04
add a comment |
$begingroup$
Faced with such a problem, I would experiment. Try different values of p, q, and n (e.g. p=3 or 5, q = 2, n =5. Look for patterns in the behavior of the sums of the fractions. Form hypotheses around the observed behavior. Prove the hypotheses. Form hypotheses around convergence. Try to prove these hypostheses. Then, if you still can't complete the problem, edit your post to show all of your work.
$endgroup$
– user2661923
Jan 3 at 17:39