Jtdylktuy

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

How should I prove this?

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

The step where you go from $lim_{ntoinfty}left(1+frac{1}{n}right)^n(n+1)- left(1+frac{1}{n-1}right)^n(n-1)$ to $lim_{ntoinfty} e(n+1)-e(n-1)$ 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
$$
nleft(frac{n+1}{n}right) - (n+1)left(frac{n}{n+1}right)
$$
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
$$
left(1 + frac{1}{x}right)^x
= eleft[1 - frac{1}{2x} + frac{11}{24x^2} - frac{7}{16x^3} + frac{2447}{5760x^4} - dotsright], ,
$$
which is derived by using a composition of the identity
$$
xlnleft(1 + frac{1}{x}right)
= 1 - frac{1}{2x} + frac{1}{3x^2} - frac{1}{4x^3} + dots
$$
with the power series expansion of $e^x$.

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

$$frac{(n+1)^{n+1}}{n^n}=(n+1)left(1+frac1nright)^n=(n+1)e^{nlogleft(1+frac1nright)}=$$$$=(n+1)e^{nleft(frac1n-frac1{2n^2}+oleft(frac1{n^2}right)right)}=(n+1)e^{1-frac1{2n}+oleft(frac1{n}right)}=e(n+1)left(1-frac1{2n}+oleft(frac1{n}right)right)$$

$$frac{n^n}{(n-1)^{n-1}}=(n-1)left(1-frac1nright)^{-n}=(n-1)e^{-nlogleft(1-frac1nright)}=$$$$=(n-1)e^{-nleft(-frac1n-frac1{2n^2}+oleft(frac1{n^2}right)right)}=(n-1)e^{1+frac1{2n}+oleft(frac1{n}right)}=e(n-1)left(1+frac1{2n}+oleft(frac1{n}right)right)$$

and then

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

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

$$a_{n+1}-a_n=frac{a_{n+1}-a_n}{(n+1)-n} to L implies frac{a_n}{n}=frac{n^{n-1}}{(n-1)^{n-1}}=frac{1}{left(1-frac1nright)^{n-1}}to L=e$$