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I have the series $\sum_{i=1}^\infty n^n/n!$ in my math textbook, and it is used as an example for the ratio test. Completing the ratio test on this problem leads to having to take the limit of:

$$\lim_{n \to \infty}(1+1/n)^n$$

the textbook says that the answer is $e$, however when I take the limit I get $$ (1+1/\infty)^\infty $$ which is just $1^\infty = 1$.

Am I missing something as to why this is not $1$? Does the variable $n$ behave different from other variables?

Sergei Levashov
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    False. $1^\infty$ is an indeterminate form, and this particular case happens to be one of the definitions of $e$. – Simply Beautiful Art Oct 15 '17 at 00:38
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    $n$ has to go to infinity in both places at the same time. Your computation is $\lim_{m\to \infty} \left( \lim_{n\to \infty} \left( 1 + \frac{1}{n} \right) \right)^m$, which is not the same as the original limit you proposed. – Xander Henderson Oct 15 '17 at 00:42
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    Also, this is a bad example for the ratio test. The term test is far simpler:$$\frac{n^n}{n!}=\frac{n\times\dots\times n}{1\times\dots\times n}>1\ne0$$ – Simply Beautiful Art Oct 15 '17 at 00:42
  • @SimplyBeautifulArt what is that expansion that you did? Im not sure how to take the original problem into a form where I can apply the term test – Sergei Levashov Oct 15 '17 at 00:48
  • @SergeiLevashov In the second comment, simply notice that $n^n=n\times n\times n\times\dots\times n$ and $n!=1\times2\times3\times\dots\times n$. Hopefully it is obvious which is bigger. – Simply Beautiful Art Oct 15 '17 at 00:50
  • Using the binomial theorem, we have for $n>2$ $$\left(1+\frac1n\right)^n=1+n,\left(\frac1n\right)+\cdots +\frac{1}{n^n}> 2$$which implies from the ratio test that the series diverges. – Mark Viola Oct 15 '17 at 00:52
  • @SimplyBeautifulArt ok, that makes sense to me, but our professor requires us to prove everything. I answered a question in a similar fashion and she said that it was not mathematical proof. – Sergei Levashov Oct 15 '17 at 00:55
  • Simply note that:$$\frac{1^1}{1!}=1\\frac{(n+1)^{n+1}}{(n+1)!}=\frac{n+1}{n+1}\frac{(n+1)^n}{n!}>\frac{n^n}{n!}>1$$and so by induction,$$\frac{n^n}{n!}>1$$and thus, it must fail the term test. – Simply Beautiful Art Oct 15 '17 at 01:01
  • How do you "take the limit" and get $(1 + 1/\infty)^\infty$? That's not even a valid expression, and taking a limit is not synonymous with just evaluating a function at a given value. – anomaly Oct 15 '17 at 02:06
  • Guessing that the limit in question should be equal to $1$ is similar to guessing that $\sqrt{2 +2}$ should be equal to $\sqrt{2}+\sqrt{2}$. It is based on an utter disregard for any rules of evaluating limits. – Paramanand Singh Oct 16 '17 at 03:26

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That's because one of the ways to define $e$ is $\lim_{n\rightarrow\infty}(1+\frac {1}{n})^n$. It's hard to see, but if you expand the polynomial using the binomial theorem and rearrange some terms and use the Taylor series of $e^x$ (which is $e^x = 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...$), and then try taking the limit, you will get $e$. This video might help.

VV6570
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Figured it out, I missed the indeterminate form $1^\infty$, so I took the natural log of both sides

ln y = ln original equation

ending up with e

Sergei Levashov
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"Without" using the definition of $e$, consider $$A_n=\left(1+\frac 1 n\right)^n\implies \log(A_n)=n\log\left(1+\frac 1 n\right)$$ Now, remember that, for small $x$, by Taylor, $$\log(1+x)=x-\frac{x^2}{2}+O\left(x^3\right)$$ Make $x=\frac 1n$ to get $$\log\left(1+\frac 1 n\right)=\frac{1}{n}-\frac{1}{2 n^2}+O\left(\frac{1}{n^3}\right)$$ $$\log(A_n)=n\log\left(1+\frac 1 n\right)=1-\frac{1}{2 n}+O\left(\frac{1}{n^2}\right)$$ Continuing with Taylor $$A=e^{\log(A)}=e-\frac{e}{2 n}+O\left(\frac{1}{n^2}\right)$$ which shows the limit and how it is approached.

Use your pocket calculator to compute $A_{10}$; you should exactly get $A_{10}=2.5937424601$ while the above approximation would give $\frac{19 }{20}e\approx 2.5823677370$. Quite close, isn't it (even so far from infinity) ?

Claude Leibovici
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