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Determine whether $a_n = (1+\frac{2}{n})^{n}$ converges or diverges. If it converges, find the limit.

So I tried to say that $a_n = (1+\frac{2}{n})^{n} \Rightarrow \ln(a_n) = n\ln(1+\frac{2}{n})$. Unfortunately I don't know what the next step is since I think that $n \rightarrow \infty$ as $n \rightarrow \infty$ and that $\ln(1+\frac{2}{n}) \rightarrow 0$ as $n\rightarrow \infty$, but somehow the solution is $e^{2}$... Can someone please help fill me in on the steps in-between? Thanks!

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Try writing $n\ln\left(1 + \frac{2}{n}\right) = \frac{\ln\left( 1 + \frac{2}{n}\right)}{\frac{1}{n}}$ and applying L'Hopital's Rule. –  JavaMan Apr 21 '11 at 18:35

3 Answers 3

up vote 8 down vote accepted

If you already know that $\displaystyle \lim_{x\to 0^+}(1+x)^{\frac{1}{x}}= e$ and that $a_n\to a$ implies $a_n^2\to a^2$, then:

$$\lim_n \left( 1+\frac{2}{n}\right)^n =\lim_n \left[ \left( 1+\frac{2}{n}\right)^{\frac{n}{2}} \right]^2=\left[\lim_n \left( 1+\frac{2}{n}\right)^{\frac{n}{2}} \right]^2 =e^2 \; .$$

Obviously, one can also write:

$$\lim_n \left( 1+\frac{2}{n}\right)^n = \lim_n \left( 1+\frac{1}{\frac{n}{2}}\right)^n =\lim_n \left[ \left( 1+\frac{1}{\frac{n}{2}}\right)^{\frac{n}{2}} \right]^2=\left[\lim_n \left( 1+\frac{1}{\frac{n}{2}}\right)^{\frac{n}{2}} \right]^2\; ,$$

and use the limit $\displaystyle \lim_{y\to +\infty} (1+\tfrac{1}{y})^y =e$.

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Thank you very much for this answer. Just one question: why is the first limit going to zero rather than infinity? –  ghshtalt Mar 29 '11 at 22:27
    
@ghshtalt: There are a lot of limits which give you $e$ and almost all of them can be obtained from the famous $\displaystyle \lim_{y\to +\infty} (1+\tfrac{1}{y})^y =e$. The limit I wrote at the beginning of my previous post is actually one of them, because it can be obtained by replacing $y$ with $\frac{1}{x}$. Moreover, that choice was convenient: in fact $\frac{2}{n} \to 0^+$, hence I had to find limit with $x\to 0^+$ which could give me $e$ if I wanted to make things work. –  Pacciu Mar 29 '11 at 22:57

We know that the derivative of $\log x$ is $\frac{1}{x}$, so apply first principle differentiation to $\log x$: $$ \begin{align} \lim_{h\to0}\frac{\log(x+h)-\log(x)}{h}&=\frac{1}{x}\\ \lim_{h\to0}\log\left(1+\frac{h}{x}\right)^{\frac{1}{h}}&=\frac{1}{x} \end{align}$$ Now replace $\frac{1}{h}$ with $n$: $$\lim_{n\to\infty}\log\left(1+\frac{1}{nx}\right)^n=\frac{1}{x}$$ Now replace $\frac{1}{x}$ by $2$: $$\begin{align} \lim_{n\to\infty}\log\left(1+\frac{2}{n}\right)^n&=2\\ \lim_{n\to\infty}e^{\log\left(1+\frac{2}{n}\right)^n}&=e^2\\ \lim_{n\to\infty}\left(1+\frac{2}{n}\right)^n&=e^2 \end{align}$$

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Why does this get a downvote? –  Vafa Khalighi Mar 30 '11 at 14:11
    
No idea. +1 from me. –  Phira Apr 21 '11 at 18:00
    
Vafa, I don't know, but both of the other answers were also downvoted. –  Jonas Meyer Apr 21 '11 at 18:02
    
Downvote was definitely unappropriate. –  Patrick Da Silva Dec 27 '11 at 5:38

You are right that $\ln (1+ \frac{2}{n}) \to 0$. However, it approaches zero only like $1/n$. Therefore, $n \ln(1+ \frac{2}{n}) = 2 + O(1/n)$ (using Taylor) and you obtain $$ a_n \to e^2.$$

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Again, unappropriate downvote. This is a very clear explanation. +1 –  Patrick Da Silva Dec 27 '11 at 5:39

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