Show that if $lim_{n to infty} x_n = 0$ if follows that $lim_{n to infty} (1+frac{x_n}{n})^n = 1$
$begingroup$
Let $(x_n)$ be a sequence in R. Show that if $lim_{n to infty} x_n = 0$ if follows that $lim_{n to infty} (1+frac{x_n}{n})^n = 1$
My idea looks like the following (using the binomial theorem):
$$(1+frac{x_n}{n})^n = sum_{k=1}^{n} {{n}choose{k}} (frac{x_n}{n})^k = sum_{k=1}^{n} frac{n!}{k! cdot (n-k)!} (frac{x_n}{n})^k=sum_{k=1}^{n} frac{n space cdot space ... space cdot space (n-k+1)}{k!} (frac{x_n}{n})^k$$
How do I proceed from here? Am I somehow supposed to split the sum up and then take the limit? Can someone help me out? Thanks in advance!
limits proof-writing limits-without-lhopital
asked Jan 2 at 12:39
John D.John D.
153
$endgroup$
|
show 1 more comment
$begingroup$
Let $(x_n)$ be a sequence in R. Show that if $lim_{n to infty} x_n = 0$ if follows that $lim_{n to infty} (1+frac{x_n}{n})^n = 1$
My idea looks like the following (using the binomial theorem):
$$(1+frac{x_n}{n})^n = sum_{k=1}^{n} {{n}choose{k}} (frac{x_n}{n})^k = sum_{k=1}^{n} frac{n!}{k! cdot (n-k)!} (frac{x_n}{n})^k=sum_{k=1}^{n} frac{n space cdot space ... space cdot space (n-k+1)}{k!} (frac{x_n}{n})^k$$
How do I proceed from here? Am I somehow supposed to split the sum up and then take the limit? Can someone help me out? Thanks in advance!
limits proof-writing limits-without-lhopital
asked Jan 2 at 12:39
John D.John D.
153
$endgroup$
$begingroup$
Are you familiar with the proposition saying that if $a_nrightarrow pm infty$, then $Big( 1+frac{1}{a_n} Big)^{a_n}rightarrow e$?
$endgroup$
– Keen-ameteur
Jan 2 at 12:49
$begingroup$
Are you sure you don't want the limit as n goes to infinity?
$endgroup$
– Zach
Jan 2 at 12:51
$begingroup$
No I'm not. I know that $lim{n to infty} (1+frac{x}{n})^n=e^x$ though. However I'm not sure how to apply that here.
$endgroup$
– John D.
Jan 2 at 12:52
$begingroup$
Oh yes as n goes to infinity.
$endgroup$
– John D.
Jan 2 at 12:52
$begingroup$
$0 le log((1+frac{x_n}{n})^n) = nlog(1+frac{x_n}{n}) le nfrac{x_n}{n} = x_n$
$endgroup$
– mathworker21
Jan 2 at 13:27
|
show 1 more comment
$begingroup$
Let $(x_n)$ be a sequence in R. Show that if $lim_{n to infty} x_n = 0$ if follows that $lim_{n to infty} (1+frac{x_n}{n})^n = 1$
My idea looks like the following (using the binomial theorem):
$$(1+frac{x_n}{n})^n = sum_{k=1}^{n} {{n}choose{k}} (frac{x_n}{n})^k = sum_{k=1}^{n} frac{n!}{k! cdot (n-k)!} (frac{x_n}{n})^k=sum_{k=1}^{n} frac{n space cdot space ... space cdot space (n-k+1)}{k!} (frac{x_n}{n})^k$$
How do I proceed from here? Am I somehow supposed to split the sum up and then take the limit? Can someone help me out? Thanks in advance!
limits proof-writing limits-without-lhopital
asked Jan 2 at 12:39
John D.John D.
153
$endgroup$
Let $(x_n)$ be a sequence in R. Show that if $lim_{n to infty} x_n = 0$ if follows that $lim_{n to infty} (1+frac{x_n}{n})^n = 1$
My idea looks like the following (using the binomial theorem):
$$(1+frac{x_n}{n})^n = sum_{k=1}^{n} {{n}choose{k}} (frac{x_n}{n})^k = sum_{k=1}^{n} frac{n!}{k! cdot (n-k)!} (frac{x_n}{n})^k=sum_{k=1}^{n} frac{n space cdot space ... space cdot space (n-k+1)}{k!} (frac{x_n}{n})^k$$
How do I proceed from here? Am I somehow supposed to split the sum up and then take the limit? Can someone help me out? Thanks in advance!
limits proof-writing limits-without-lhopital
limits proof-writing limits-without-lhopital
asked Jan 2 at 12:39
John D.John D.
153
asked Jan 2 at 12:39
John D.John D.
153
edited Jan 2 at 12:53
John D.
asked Jan 2 at 12:39
John D.John D.
153
asked Jan 2 at 12:39
John D.John D.
153
asked Jan 2 at 12:39
John D.John D.
153
153
$begingroup$
Are you familiar with the proposition saying that if $a_nrightarrow pm infty$, then $Big( 1+frac{1}{a_n} Big)^{a_n}rightarrow e$?
$endgroup$
– Keen-ameteur
Jan 2 at 12:49
$begingroup$
Are you sure you don't want the limit as n goes to infinity?
$endgroup$
– Zach
Jan 2 at 12:51
$begingroup$
No I'm not. I know that $lim{n to infty} (1+frac{x}{n})^n=e^x$ though. However I'm not sure how to apply that here.
$endgroup$
– John D.
Jan 2 at 12:52
$begingroup$
Oh yes as n goes to infinity.
$endgroup$
– John D.
Jan 2 at 12:52
$begingroup$
$0 le log((1+frac{x_n}{n})^n) = nlog(1+frac{x_n}{n}) le nfrac{x_n}{n} = x_n$
$endgroup$
– mathworker21
Jan 2 at 13:27
|
show 1 more comment
$begingroup$
Are you familiar with the proposition saying that if $a_nrightarrow pm infty$, then $Big( 1+frac{1}{a_n} Big)^{a_n}rightarrow e$?
$endgroup$
– Keen-ameteur
Jan 2 at 12:49
$begingroup$
Are you sure you don't want the limit as n goes to infinity?
$endgroup$
– Zach
Jan 2 at 12:51
$begingroup$
No I'm not. I know that $lim{n to infty} (1+frac{x}{n})^n=e^x$ though. However I'm not sure how to apply that here.
$endgroup$
– John D.
Jan 2 at 12:52
$begingroup$
Oh yes as n goes to infinity.
$endgroup$
– John D.
Jan 2 at 12:52
$begingroup$
$0 le log((1+frac{x_n}{n})^n) = nlog(1+frac{x_n}{n}) le nfrac{x_n}{n} = x_n$
$endgroup$
– mathworker21
Jan 2 at 13:27
$begingroup$
Are you familiar with the proposition saying that if $a_nrightarrow pm infty$, then $Big( 1+frac{1}{a_n} Big)^{a_n}rightarrow e$?
$endgroup$
– Keen-ameteur
Jan 2 at 12:49
$begingroup$
Are you familiar with the proposition saying that if $a_nrightarrow pm infty$, then $Big( 1+frac{1}{a_n} Big)^{a_n}rightarrow e$?
$endgroup$
– Keen-ameteur
Jan 2 at 12:49
$begingroup$
Are you sure you don't want the limit as n goes to infinity?
$endgroup$
– Zach
Jan 2 at 12:51
$begingroup$
Are you sure you don't want the limit as n goes to infinity?
$endgroup$
– Zach
Jan 2 at 12:51
$begingroup$
No I'm not. I know that $lim{n to infty} (1+frac{x}{n})^n=e^x$ though. However I'm not sure how to apply that here.
$endgroup$
– John D.
Jan 2 at 12:52
$begingroup$
No I'm not. I know that $lim{n to infty} (1+frac{x}{n})^n=e^x$ though. However I'm not sure how to apply that here.
$endgroup$
– John D.
Jan 2 at 12:52
$begingroup$
Oh yes as n goes to infinity.
$endgroup$
– John D.
Jan 2 at 12:52
$begingroup$
Oh yes as n goes to infinity.
$endgroup$
– John D.
Jan 2 at 12:52
$begingroup$
$0 le log((1+frac{x_n}{n})^n) = nlog(1+frac{x_n}{n}) le nfrac{x_n}{n} = x_n$
$endgroup$
– mathworker21
Jan 2 at 13:27
$begingroup$
$0 le log((1+frac{x_n}{n})^n) = nlog(1+frac{x_n}{n}) le nfrac{x_n}{n} = x_n$
$endgroup$
– mathworker21
Jan 2 at 13:27
|
show 1 more comment
1 Answer
1
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$begingroup$
For $n$ sufficiently large $-epsilon <x_n <epsilon$ so $(1-frac {epsilon} n)^{n} leq (1-frac {x_n} n)^{n} leq (1-frac {epsilon} n)^{n}$. Use squeeze theorem, the fact that $(1+frac x n)^{n} to e^{x}$ for any real number $x$, and then observe that $e^{-epsilon}$ and $e^{epsilon}$ both tend to $1$ as $ epsilon to 0$.
answered Jan 2 at 13:01
Kavi Rama MurthyKavi Rama Murthy
63.7k42464
$endgroup$
$begingroup$
Thank you very much!
$endgroup$
– John D.
Jan 2 at 13:14
add a comment |
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1 Answer
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1 Answer
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$begingroup$
For $n$ sufficiently large $-epsilon <x_n <epsilon$ so $(1-frac {epsilon} n)^{n} leq (1-frac {x_n} n)^{n} leq (1-frac {epsilon} n)^{n}$. Use squeeze theorem, the fact that $(1+frac x n)^{n} to e^{x}$ for any real number $x$, and then observe that $e^{-epsilon}$ and $e^{epsilon}$ both tend to $1$ as $ epsilon to 0$.
answered Jan 2 at 13:01
Kavi Rama MurthyKavi Rama Murthy
63.7k42464
$endgroup$
$begingroup$
Thank you very much!
$endgroup$
– John D.
Jan 2 at 13:14
add a comment |
$begingroup$
For $n$ sufficiently large $-epsilon <x_n <epsilon$ so $(1-frac {epsilon} n)^{n} leq (1-frac {x_n} n)^{n} leq (1-frac {epsilon} n)^{n}$. Use squeeze theorem, the fact that $(1+frac x n)^{n} to e^{x}$ for any real number $x$, and then observe that $e^{-epsilon}$ and $e^{epsilon}$ both tend to $1$ as $ epsilon to 0$.
answered Jan 2 at 13:01
Kavi Rama MurthyKavi Rama Murthy
63.7k42464
$endgroup$
$begingroup$
Thank you very much!
$endgroup$
– John D.
Jan 2 at 13:14
add a comment |
$begingroup$
For $n$ sufficiently large $-epsilon <x_n <epsilon$ so $(1-frac {epsilon} n)^{n} leq (1-frac {x_n} n)^{n} leq (1-frac {epsilon} n)^{n}$. Use squeeze theorem, the fact that $(1+frac x n)^{n} to e^{x}$ for any real number $x$, and then observe that $e^{-epsilon}$ and $e^{epsilon}$ both tend to $1$ as $ epsilon to 0$.
answered Jan 2 at 13:01
Kavi Rama MurthyKavi Rama Murthy
63.7k42464
$endgroup$
For $n$ sufficiently large $-epsilon <x_n <epsilon$ so $(1-frac {epsilon} n)^{n} leq (1-frac {x_n} n)^{n} leq (1-frac {epsilon} n)^{n}$. Use squeeze theorem, the fact that $(1+frac x n)^{n} to e^{x}$ for any real number $x$, and then observe that $e^{-epsilon}$ and $e^{epsilon}$ both tend to $1$ as $ epsilon to 0$.
answered Jan 2 at 13:01
Kavi Rama MurthyKavi Rama Murthy
63.7k42464
answered Jan 2 at 13:01
Kavi Rama MurthyKavi Rama Murthy
63.7k42464
answered Jan 2 at 13:01
Kavi Rama MurthyKavi Rama Murthy
63.7k42464
answered Jan 2 at 13:01
Kavi Rama MurthyKavi Rama Murthy
63.7k42464
63.7k42464
$begingroup$
Thank you very much!
$endgroup$
– John D.
Jan 2 at 13:14
add a comment |
$begingroup$
Thank you very much!
$endgroup$
– John D.
Jan 2 at 13:14
$begingroup$
Thank you very much!
$endgroup$
– John D.
Jan 2 at 13:14
$begingroup$
Thank you very much!
$endgroup$
– John D.
Jan 2 at 13:14
add a comment |
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$begingroup$
Are you familiar with the proposition saying that if $a_nrightarrow pm infty$, then $Big( 1+frac{1}{a_n} Big)^{a_n}rightarrow e$?
$endgroup$
– Keen-ameteur
Jan 2 at 12:49
$begingroup$
Are you sure you don't want the limit as n goes to infinity?
$endgroup$
– Zach
Jan 2 at 12:51
$begingroup$
No I'm not. I know that $lim{n to infty} (1+frac{x}{n})^n=e^x$ though. However I'm not sure how to apply that here.
$endgroup$
– John D.
Jan 2 at 12:52
$begingroup$
Oh yes as n goes to infinity.
$endgroup$
– John D.
Jan 2 at 12:52
$begingroup$
$0 le log((1+frac{x_n}{n})^n) = nlog(1+frac{x_n}{n}) le nfrac{x_n}{n} = x_n$
$endgroup$
– mathworker21
Jan 2 at 13:27