I believe the answer is just zero, is it not? Since we have $(1+1/n)^{n}= e$ Or am I missing something?

No. It's not 0. That's what I believed and tried to prove during the contest If you want, I can post the solution...

edit: Ohps, read the problem wrong. Well, this is certainly not more difficult the way it is phrased. Taking logarithms we get $a_{n}= \frac{1}{\ln\left(1+1/n\right)}-n$ and by taylor expansions we get $\lim_{n}a_{n}= 1/2$

Also the idea that $\lim_{n \to \infty}a_{n}= \frac{1}{2}$ can be observed from the Stirling formula: (1) $n! \equiv \sqrt{2\pi}\cdot \frac{n^{n+\frac{1}{2}}}{e^{n}}$ $(n+1)! \equiv \sqrt{2\pi}\cdot \frac{(n+1)^{n+1+\frac{1}{2}}}{e^{n+1}}$ or (2) $n! \equiv \sqrt{2\pi}\cdot \frac{(n+1)^{n+\frac{1}{2}}}{e^{n+1}}$ Comparing them $e\equiv \left (\frac{n+1}{n}\right)^{n+\frac{1}{2}}$

prowler wrote: Also the idea that $\lim_{n \to \infty}a_{n}= \frac{1}{2}$ can be observed from the Stirling formula: (1) $n! \equiv \sqrt{2\pi}\cdot \frac{n^{n+\frac{1}{2}}}{e^{n}}$ $(n+1)! \equiv \sqrt{2\pi}\cdot \frac{(n+1)^{n+1+\frac{1}{2}}}{e^{n+1}}$ or (2) $n! \equiv \sqrt{2\pi}\cdot \frac{(n+1)^{n+\frac{1}{2}}}{e^{n+1}}$ Comparing them $e\equiv \left (\frac{n+1}{n}\right)^{n+\frac{1}{2}}$ Yeah.. prowler Do you think I don't who proposed this problem? Good luck!

take logarithm $n$+$a_{n}$=$\frac{1}{{\ln \left({1+\frac{1}{n}}\right)}}$ $n$+$a_{n}$=$\frac{1}{\frac{1}{n}-\frac{1}{2n^{2}}+o(\frac{1}{n^{2}})}$ $n$+$a_{n}$=$n\frac{1}{1-\frac{1}{2n}+o(\frac{1}{n})}$ $n$+$a_{n}$=$n(1+\frac{1}{2n}+o(\frac{1}{n}))$ $n$+$a_{n}$=$n+\frac{1}{2}+o(1)$ $a_{n}$=$\frac{1}{2}+o(1)$

I don't think the limit equal $ \frac{1}{2}$ becaese we have $ a_n<0,\forall n$. Indeed $ \frac{1}{\ln\left(1+\frac{1}{n}\right)}-n<0$ is equavalent to $ 1<\ln\left(1+\frac{1}{n}\right)^n$. But that is clear because we have $ 2<\left(1+\frac{1}{n}\right)^n<3$. The anwser is 0

I'm afraid I can't agree. $ \left(1+\frac{1}{n}\right)^n$ is monotone increasing with limit $ e$, hence $ \left(1+\frac{1}{n}\right)^n<e$, so $ \ln\left(1+\frac{1}{n}\right)^n<1$.

Yes, this is my mistake.