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I learnt on Wolfram MathWorld that $$\lim_{n\rightarrow\infty}\left[\frac{(n+1)^{n+1}}{n^n}-\frac{n^n}{(n-1)^{n-1}}\right] = e$$

How should I prove this?

Attempt: $$\begin{align} \lim_{n\rightarrow\infty}\left[\frac{(n+1)^{n+1}}{n^n}-\frac{n^n}{(n-1)^{n-1}}\right] &= \lim_{n\rightarrow\infty}\left[\left(1+\frac{1}{n}\right)^n(n+1)-\left(1+\frac{1}{n-1}\right)^n(n-1)\right]\\ &=\lim_{n\rightarrow\infty}e(n+1)-e(n-1)\\ &= 2e \end{align}$$ Why did I get a $2e$? Where did I do wrong? Isn't $$\lim_{n\rightarrow\infty}\left(1+\frac{1}{n-1}\right)^n=e?$$

Community
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Larry
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    $$ \left(1+\frac{1}{n-1} \right )^{n} = \left(1+\frac{1}{n-1} \right )^{n-1} \cdot \left(1+\frac{1}{n-1} \right ) $$ – Vladislav Kharlamov Dec 08 '18 at 01:05
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    those $$ e(n+1)-e \cdot \left (1+\frac{1}{n-1} \right )(n-1) = e(n+1)-e(n-1+1) = e(n+1)-en = e $$ – Vladislav Kharlamov Dec 08 '18 at 01:08
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    The exponent on the second term must match the denominator (you have an exponent of $n$, when the denominator is $n-1$). – Michael Burr Dec 08 '18 at 01:12
  • Thank you guys, your answers are really helpful. You could change your comments to answers if you don't mind. – Larry Dec 08 '18 at 01:30
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    The step $$\lim_{n\rightarrow\infty}\left(\left(1+\frac{1}{n}\right)^n(n+1)-\left(1+\frac{1}{n-1}\right)^n(n-1)\right)=\lim_{n\rightarrow\infty}(e(n+1)-e(n-1))$$ is erroneous. – Abraham Zhang Dec 08 '18 at 01:33
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    Your attempt has a flaw which is very common. See this answer for details: https://math.stackexchange.com/a/1783818/72031 – Paramanand Singh Dec 08 '18 at 02:09
  • @Paramanand: Thanks for pointing that out. I did not realize that someone had already asked the similar question. – Larry Dec 08 '18 at 03:03

3 Answers3

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The step where you go from $\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n(n+1)- \left(1+\frac{1}{n-1}\right)^n(n-1)$ to $\lim_{n\to\infty} e(n+1)-e(n-1)$ is invalid. You can't evaluate just some pieces of the limit in terms of $n$. Formally, you're starting with an indeterminate form of type $\infty - \infty$.

As a simpler example as to why your reasoning doesn't work, consider the indeterminate form $(n+1)-n$, which is identically equal to $1$. But it can be rewritten to be of the form $$ n\left(\frac{n+1}{n}\right) - (n+1)\left(\frac{n}{n+1}\right) $$ Now both $\left(\frac{n+1}{n}\right)$ and $\left(\frac{n}{n+1}\right)$ converge to $1$, but the limit of $n(1) - (n+1)(1)$ is now $-1$ instead of $1$.

As to how to derive the identity, the Mathworld link suggests that a reasonable approach would be to start with the expression $$ \left(1 + \frac{1}{x}\right)^x = e\left[1 - \frac{1}{2x} + \frac{11}{24x^2} - \frac{7}{16x^3} + \frac{2447}{5760x^4} - \dots\right]\, , $$ which is derived by using a composition of the identity $$ x\ln\left(1 + \frac{1}{x}\right) = 1 - \frac{1}{2x} + \frac{1}{3x^2} - \frac{1}{4x^3} + \dots $$ with the power series expansion of $e^x$.

Rolf Hoyer
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By Taylor's espansion as $x \to 0$

  • $\log(1+x)=x-\frac12x^2+o(x^2)$
  • $e^x=1+x+o(x)$

we have that

$$\frac{(n+1)^{n+1}}{n^n}=(n+1)\left(1+\frac1n\right)^n=(n+1)e^{n\log\left(1+\frac1n\right)}=$$$$=(n+1)e^{n\left(\frac1n-\frac1{2n^2}+o\left(\frac1{n^2}\right)\right)}=(n+1)e^{1-\frac1{2n}+o\left(\frac1{n}\right)}=e(n+1)\left(1-\frac1{2n}+o\left(\frac1{n}\right)\right)$$

$$\frac{n^n}{(n-1)^{n-1}}=(n-1)\left(1-\frac1n\right)^{-n}=(n-1)e^{-n\log\left(1-\frac1n\right)}=$$$$=(n-1)e^{-n\left(-\frac1n-\frac1{2n^2}+o\left(\frac1{n^2}\right)\right)}=(n-1)e^{1+\frac1{2n}+o\left(\frac1{n}\right)}=e(n-1)\left(1+\frac1{2n}+o\left(\frac1{n}\right)\right)$$

and then

$$\frac{(n+1)^{n+1}}{n^n}-\frac{n^n}{(n-1)^{n-1}}=en+e-\frac e 2-(en-e+\frac e 2)+o(1)=e+o(1)\to e$$

user
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Assuming that the limit exists (which should be proved) we can easily see that it needs to be equal to $e$, indeed by Stolz-Cesaro with $a_n=\frac{n^n}{(n-1)^{n-1}}$ we have

$$a_{n+1}-a_n=\frac{a_{n+1}-a_n}{(n+1)-n} \to L \implies \frac{a_n}{n}=\frac{n^{n-1}}{(n-1)^{n-1}}=\frac{1}{\left(1-\frac1n\right)^{n-1}}\to L=e$$

user
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