Find the limit of the sequence $a_n=\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}.$
The limit is supposed to be $e^{-1}$ and there are a couple of posts in MSE proving it. But here is a proof I encountered showing the limit is $1.$ Please tell me what's going wrong here.

Proof: We can write $a_n=\sqrt[n]{n!}(b_n-1),$ where $b_n=\sqrt[n+1]{(n+1)!}/\sqrt[n]{n!}.$ Hence $$a_n=\frac{\sqrt[n]{n!}}{n}.\frac{b_n-1}{\ln{b_n}}.\ln{b_n^n.}$$ But $\lim \limits_{n\to \infty}\sqrt[n]{n!}/n=e^{-1},$ so $b_n\to1$ as $n\to \infty.$ On the other hand, $$\lim \limits_{n\to \infty}b_n^n=\lim \limits_{n\to \infty}\frac{(n+1)!}{n!}.\frac{1}{\sqrt[n+1]{(n+1)!}}=e.$$ Also $$\lim \limits_{n\to \infty}\frac{b_n-1}{\ln{b_n}}=\lim \limits_{n\to \infty}\frac{1}{\ln{[1+(b_n-1)]^{1/(b_n-1)}}}=\frac{1}{\ln e}=1$$

And the author goes on to say that the limit is $e^{-1}.$ But doesn't it say that the limit is $1?$ I must have overlooked something so obvious.


$$\begin{align} \lim_{n\to \infty}a_n&=\color{blue}{\lim_{n\to \infty}\left(\frac{\sqrt[n]{n!}}{n}\right)}\color{red}{\lim_{n\to \infty}\left(\frac{b_n-1}{\log(b_n)}\right)}\color{green}{\lim_{n\to \infty}\left(\log(b_n^n)\right)}\\\\ &=\color{blue}{(e^{-1})}\color{red}{(1)}\color{green}{(\log(e))}\\\\ &=\color{blue}{(e^{-1})}\color{red}{(1)}\color{green}{(1)}\\\\ &=e^{-1} \end{align}$$

  • $\begingroup$ OMG!!!!! Here I sit realizing how many hours I wasted. Thanks a lot. $\endgroup$ – Bijesh K.S Nov 18 '16 at 16:45
  • 2
    $\begingroup$ You're welcome! My pleasure. -Mark $\endgroup$ – Mark Viola Nov 18 '16 at 16:46

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