How to find $lim_{ntoinfty}(pa_n + qb_n)^n$ with $p + q = 1$?
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Suppose $a_n$ and $b_n$ are sequences of positive numbers, such that
$$
lim_{ntoinfty}a_n^n = a,quad lim_{ntoinfty}b_n^n = b,qquad a,bin (0, infty).
$$
Find limit
$$
lim_{ntoinfty}(pa_n + qb_n)^n,
$$
where $p, q$ are nonnegative numbers such that $p + q = 1$.
calculus sequences-and-series limits
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up vote
5
down vote
favorite
Suppose $a_n$ and $b_n$ are sequences of positive numbers, such that
$$
lim_{ntoinfty}a_n^n = a,quad lim_{ntoinfty}b_n^n = b,qquad a,bin (0, infty).
$$
Find limit
$$
lim_{ntoinfty}(pa_n + qb_n)^n,
$$
where $p, q$ are nonnegative numbers such that $p + q = 1$.
calculus sequences-and-series limits
When $a>b$, consider $$(pa_n + qb_n)^n=a^nleft(p+qcdot frac{b_n}{a_n}right)^n.$$
– user587192
yesterday
Very similar: math.stackexchange.com/q/2995279/587192
– user587192
yesterday
add a comment |
up vote
5
down vote
favorite
up vote
5
down vote
favorite
Suppose $a_n$ and $b_n$ are sequences of positive numbers, such that
$$
lim_{ntoinfty}a_n^n = a,quad lim_{ntoinfty}b_n^n = b,qquad a,bin (0, infty).
$$
Find limit
$$
lim_{ntoinfty}(pa_n + qb_n)^n,
$$
where $p, q$ are nonnegative numbers such that $p + q = 1$.
calculus sequences-and-series limits
Suppose $a_n$ and $b_n$ are sequences of positive numbers, such that
$$
lim_{ntoinfty}a_n^n = a,quad lim_{ntoinfty}b_n^n = b,qquad a,bin (0, infty).
$$
Find limit
$$
lim_{ntoinfty}(pa_n + qb_n)^n,
$$
where $p, q$ are nonnegative numbers such that $p + q = 1$.
calculus sequences-and-series limits
calculus sequences-and-series limits
edited yesterday
user587192
1,01310
1,01310
asked yesterday
J. Abraham
464313
464313
When $a>b$, consider $$(pa_n + qb_n)^n=a^nleft(p+qcdot frac{b_n}{a_n}right)^n.$$
– user587192
yesterday
Very similar: math.stackexchange.com/q/2995279/587192
– user587192
yesterday
add a comment |
When $a>b$, consider $$(pa_n + qb_n)^n=a^nleft(p+qcdot frac{b_n}{a_n}right)^n.$$
– user587192
yesterday
Very similar: math.stackexchange.com/q/2995279/587192
– user587192
yesterday
When $a>b$, consider $$(pa_n + qb_n)^n=a^nleft(p+qcdot frac{b_n}{a_n}right)^n.$$
– user587192
yesterday
When $a>b$, consider $$(pa_n + qb_n)^n=a^nleft(p+qcdot frac{b_n}{a_n}right)^n.$$
– user587192
yesterday
Very similar: math.stackexchange.com/q/2995279/587192
– user587192
yesterday
Very similar: math.stackexchange.com/q/2995279/587192
– user587192
yesterday
add a comment |
2 Answers
2
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oldest
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up vote
2
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Write $a_n^n = e^{alpha_n}$ and $b_n^n = e^{beta_n}$. Then $alpha_n to log a$ , $beta_n to log b$ and
$$ a_n = 1 + frac{alpha_n}{n} + mathcal{O}left(frac{1}{n^2}right), qquad b_n = 1 + frac{beta_n}{n} + mathcal{O}left(frac{1}{n^2}right). $$
Therefore
$$ (p a_n + q b_n)^n = left( 1 + frac{p alpha_n + q beta_n}{n} + mathcal{O}left(frac{1}{n^2}right) right)^n xrightarrow[ntoinfty]{} e^{p log a + q log b} = a^p b^q. $$
How did you derive a formula for $a_n$ and $b_n$? Is there a way to do this, without asymptotics?
– user4201961
21 hours ago
@user4201961: this is nothing but Taylor series for $e^x$. Just note that $a_n=e^{alpha_n/n} $.
– Paramanand Singh
17 hours ago
add a comment |
up vote
1
down vote
We use the following lemmas:
Lemma 1: Let ${a_n} $ be a sequence such that $a_nto a>0$. Then $n(a_n^{1/n}-1)tolog a$ (prove it).
Lemma 2 : Let ${a_n} $ be a sequence such that $n(a_n-1)to 0$. Then $a_n^nto 1$.
Consider the sequence ${x_n} $ defined by $$x_n=frac{pa_n+qb_n} {a_n^pb_n^q} $$ and then we have $$n(x_n-1)=frac{n((a_n^{np}
b_n^{nq}) ^{1/n}-1)-pn((a_n^n)^{1/n}-1)-qn((b_n^n)^{1/n}-1)}{a_n^pb_n^q}$$ By lemma $1$ numerator of the above expression tends to $$log(a^pb^q) - plog a-qlog b=0$$ and since $a_nto 1,b_nto 1$ (prove this) the denominator tends to $1$ so that $n(x_n-1)to 0$. Thus by lemma 2 the sequence $x_n^nto 1$ and thus the desired limit is equal to $a^pb^q$.
Alternatively we can use lemma 1 and the following
Lemma 3: Let ${a_n} $ be a sequence such that $a_nto a$ then $$left(1+frac{a_n}{n}right)^nto e^a$$
We can write $$pa_n+qb_n=1+frac{x_n}{n}$$ where $$x_n=pn((a_n^{n})^{1/n}-1)+qn((b_n^{n})^{1/n}-1)$$ and by lemma 1 $$x_nto plog a+qlog b=log(a^pb^q) $$ and therefore by lemma 3 we have $$(pa_n+qb_n)^n=left(1+frac{x_n}{n}right)^nto exp(log(a^pb^q)) =a^pb^q$$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Write $a_n^n = e^{alpha_n}$ and $b_n^n = e^{beta_n}$. Then $alpha_n to log a$ , $beta_n to log b$ and
$$ a_n = 1 + frac{alpha_n}{n} + mathcal{O}left(frac{1}{n^2}right), qquad b_n = 1 + frac{beta_n}{n} + mathcal{O}left(frac{1}{n^2}right). $$
Therefore
$$ (p a_n + q b_n)^n = left( 1 + frac{p alpha_n + q beta_n}{n} + mathcal{O}left(frac{1}{n^2}right) right)^n xrightarrow[ntoinfty]{} e^{p log a + q log b} = a^p b^q. $$
How did you derive a formula for $a_n$ and $b_n$? Is there a way to do this, without asymptotics?
– user4201961
21 hours ago
@user4201961: this is nothing but Taylor series for $e^x$. Just note that $a_n=e^{alpha_n/n} $.
– Paramanand Singh
17 hours ago
add a comment |
up vote
2
down vote
Write $a_n^n = e^{alpha_n}$ and $b_n^n = e^{beta_n}$. Then $alpha_n to log a$ , $beta_n to log b$ and
$$ a_n = 1 + frac{alpha_n}{n} + mathcal{O}left(frac{1}{n^2}right), qquad b_n = 1 + frac{beta_n}{n} + mathcal{O}left(frac{1}{n^2}right). $$
Therefore
$$ (p a_n + q b_n)^n = left( 1 + frac{p alpha_n + q beta_n}{n} + mathcal{O}left(frac{1}{n^2}right) right)^n xrightarrow[ntoinfty]{} e^{p log a + q log b} = a^p b^q. $$
How did you derive a formula for $a_n$ and $b_n$? Is there a way to do this, without asymptotics?
– user4201961
21 hours ago
@user4201961: this is nothing but Taylor series for $e^x$. Just note that $a_n=e^{alpha_n/n} $.
– Paramanand Singh
17 hours ago
add a comment |
up vote
2
down vote
up vote
2
down vote
Write $a_n^n = e^{alpha_n}$ and $b_n^n = e^{beta_n}$. Then $alpha_n to log a$ , $beta_n to log b$ and
$$ a_n = 1 + frac{alpha_n}{n} + mathcal{O}left(frac{1}{n^2}right), qquad b_n = 1 + frac{beta_n}{n} + mathcal{O}left(frac{1}{n^2}right). $$
Therefore
$$ (p a_n + q b_n)^n = left( 1 + frac{p alpha_n + q beta_n}{n} + mathcal{O}left(frac{1}{n^2}right) right)^n xrightarrow[ntoinfty]{} e^{p log a + q log b} = a^p b^q. $$
Write $a_n^n = e^{alpha_n}$ and $b_n^n = e^{beta_n}$. Then $alpha_n to log a$ , $beta_n to log b$ and
$$ a_n = 1 + frac{alpha_n}{n} + mathcal{O}left(frac{1}{n^2}right), qquad b_n = 1 + frac{beta_n}{n} + mathcal{O}left(frac{1}{n^2}right). $$
Therefore
$$ (p a_n + q b_n)^n = left( 1 + frac{p alpha_n + q beta_n}{n} + mathcal{O}left(frac{1}{n^2}right) right)^n xrightarrow[ntoinfty]{} e^{p log a + q log b} = a^p b^q. $$
answered yesterday
Sangchul Lee
89.2k12161261
89.2k12161261
How did you derive a formula for $a_n$ and $b_n$? Is there a way to do this, without asymptotics?
– user4201961
21 hours ago
@user4201961: this is nothing but Taylor series for $e^x$. Just note that $a_n=e^{alpha_n/n} $.
– Paramanand Singh
17 hours ago
add a comment |
How did you derive a formula for $a_n$ and $b_n$? Is there a way to do this, without asymptotics?
– user4201961
21 hours ago
@user4201961: this is nothing but Taylor series for $e^x$. Just note that $a_n=e^{alpha_n/n} $.
– Paramanand Singh
17 hours ago
How did you derive a formula for $a_n$ and $b_n$? Is there a way to do this, without asymptotics?
– user4201961
21 hours ago
How did you derive a formula for $a_n$ and $b_n$? Is there a way to do this, without asymptotics?
– user4201961
21 hours ago
@user4201961: this is nothing but Taylor series for $e^x$. Just note that $a_n=e^{alpha_n/n} $.
– Paramanand Singh
17 hours ago
@user4201961: this is nothing but Taylor series for $e^x$. Just note that $a_n=e^{alpha_n/n} $.
– Paramanand Singh
17 hours ago
add a comment |
up vote
1
down vote
We use the following lemmas:
Lemma 1: Let ${a_n} $ be a sequence such that $a_nto a>0$. Then $n(a_n^{1/n}-1)tolog a$ (prove it).
Lemma 2 : Let ${a_n} $ be a sequence such that $n(a_n-1)to 0$. Then $a_n^nto 1$.
Consider the sequence ${x_n} $ defined by $$x_n=frac{pa_n+qb_n} {a_n^pb_n^q} $$ and then we have $$n(x_n-1)=frac{n((a_n^{np}
b_n^{nq}) ^{1/n}-1)-pn((a_n^n)^{1/n}-1)-qn((b_n^n)^{1/n}-1)}{a_n^pb_n^q}$$ By lemma $1$ numerator of the above expression tends to $$log(a^pb^q) - plog a-qlog b=0$$ and since $a_nto 1,b_nto 1$ (prove this) the denominator tends to $1$ so that $n(x_n-1)to 0$. Thus by lemma 2 the sequence $x_n^nto 1$ and thus the desired limit is equal to $a^pb^q$.
Alternatively we can use lemma 1 and the following
Lemma 3: Let ${a_n} $ be a sequence such that $a_nto a$ then $$left(1+frac{a_n}{n}right)^nto e^a$$
We can write $$pa_n+qb_n=1+frac{x_n}{n}$$ where $$x_n=pn((a_n^{n})^{1/n}-1)+qn((b_n^{n})^{1/n}-1)$$ and by lemma 1 $$x_nto plog a+qlog b=log(a^pb^q) $$ and therefore by lemma 3 we have $$(pa_n+qb_n)^n=left(1+frac{x_n}{n}right)^nto exp(log(a^pb^q)) =a^pb^q$$
add a comment |
up vote
1
down vote
We use the following lemmas:
Lemma 1: Let ${a_n} $ be a sequence such that $a_nto a>0$. Then $n(a_n^{1/n}-1)tolog a$ (prove it).
Lemma 2 : Let ${a_n} $ be a sequence such that $n(a_n-1)to 0$. Then $a_n^nto 1$.
Consider the sequence ${x_n} $ defined by $$x_n=frac{pa_n+qb_n} {a_n^pb_n^q} $$ and then we have $$n(x_n-1)=frac{n((a_n^{np}
b_n^{nq}) ^{1/n}-1)-pn((a_n^n)^{1/n}-1)-qn((b_n^n)^{1/n}-1)}{a_n^pb_n^q}$$ By lemma $1$ numerator of the above expression tends to $$log(a^pb^q) - plog a-qlog b=0$$ and since $a_nto 1,b_nto 1$ (prove this) the denominator tends to $1$ so that $n(x_n-1)to 0$. Thus by lemma 2 the sequence $x_n^nto 1$ and thus the desired limit is equal to $a^pb^q$.
Alternatively we can use lemma 1 and the following
Lemma 3: Let ${a_n} $ be a sequence such that $a_nto a$ then $$left(1+frac{a_n}{n}right)^nto e^a$$
We can write $$pa_n+qb_n=1+frac{x_n}{n}$$ where $$x_n=pn((a_n^{n})^{1/n}-1)+qn((b_n^{n})^{1/n}-1)$$ and by lemma 1 $$x_nto plog a+qlog b=log(a^pb^q) $$ and therefore by lemma 3 we have $$(pa_n+qb_n)^n=left(1+frac{x_n}{n}right)^nto exp(log(a^pb^q)) =a^pb^q$$
add a comment |
up vote
1
down vote
up vote
1
down vote
We use the following lemmas:
Lemma 1: Let ${a_n} $ be a sequence such that $a_nto a>0$. Then $n(a_n^{1/n}-1)tolog a$ (prove it).
Lemma 2 : Let ${a_n} $ be a sequence such that $n(a_n-1)to 0$. Then $a_n^nto 1$.
Consider the sequence ${x_n} $ defined by $$x_n=frac{pa_n+qb_n} {a_n^pb_n^q} $$ and then we have $$n(x_n-1)=frac{n((a_n^{np}
b_n^{nq}) ^{1/n}-1)-pn((a_n^n)^{1/n}-1)-qn((b_n^n)^{1/n}-1)}{a_n^pb_n^q}$$ By lemma $1$ numerator of the above expression tends to $$log(a^pb^q) - plog a-qlog b=0$$ and since $a_nto 1,b_nto 1$ (prove this) the denominator tends to $1$ so that $n(x_n-1)to 0$. Thus by lemma 2 the sequence $x_n^nto 1$ and thus the desired limit is equal to $a^pb^q$.
Alternatively we can use lemma 1 and the following
Lemma 3: Let ${a_n} $ be a sequence such that $a_nto a$ then $$left(1+frac{a_n}{n}right)^nto e^a$$
We can write $$pa_n+qb_n=1+frac{x_n}{n}$$ where $$x_n=pn((a_n^{n})^{1/n}-1)+qn((b_n^{n})^{1/n}-1)$$ and by lemma 1 $$x_nto plog a+qlog b=log(a^pb^q) $$ and therefore by lemma 3 we have $$(pa_n+qb_n)^n=left(1+frac{x_n}{n}right)^nto exp(log(a^pb^q)) =a^pb^q$$
We use the following lemmas:
Lemma 1: Let ${a_n} $ be a sequence such that $a_nto a>0$. Then $n(a_n^{1/n}-1)tolog a$ (prove it).
Lemma 2 : Let ${a_n} $ be a sequence such that $n(a_n-1)to 0$. Then $a_n^nto 1$.
Consider the sequence ${x_n} $ defined by $$x_n=frac{pa_n+qb_n} {a_n^pb_n^q} $$ and then we have $$n(x_n-1)=frac{n((a_n^{np}
b_n^{nq}) ^{1/n}-1)-pn((a_n^n)^{1/n}-1)-qn((b_n^n)^{1/n}-1)}{a_n^pb_n^q}$$ By lemma $1$ numerator of the above expression tends to $$log(a^pb^q) - plog a-qlog b=0$$ and since $a_nto 1,b_nto 1$ (prove this) the denominator tends to $1$ so that $n(x_n-1)to 0$. Thus by lemma 2 the sequence $x_n^nto 1$ and thus the desired limit is equal to $a^pb^q$.
Alternatively we can use lemma 1 and the following
Lemma 3: Let ${a_n} $ be a sequence such that $a_nto a$ then $$left(1+frac{a_n}{n}right)^nto e^a$$
We can write $$pa_n+qb_n=1+frac{x_n}{n}$$ where $$x_n=pn((a_n^{n})^{1/n}-1)+qn((b_n^{n})^{1/n}-1)$$ and by lemma 1 $$x_nto plog a+qlog b=log(a^pb^q) $$ and therefore by lemma 3 we have $$(pa_n+qb_n)^n=left(1+frac{x_n}{n}right)^nto exp(log(a^pb^q)) =a^pb^q$$
edited 16 hours ago
answered yesterday
Paramanand Singh
47.8k555152
47.8k555152
add a comment |
add a comment |
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When $a>b$, consider $$(pa_n + qb_n)^n=a^nleft(p+qcdot frac{b_n}{a_n}right)^n.$$
– user587192
yesterday
Very similar: math.stackexchange.com/q/2995279/587192
– user587192
yesterday