# Prove $\displaystyle n\log\left(1+\frac{1}{n}\right) - \log\left(1+\frac{1}{n+1}\right) < \log\left(1+\frac{1}{n+1}\right)$

I'm trying to prove the above inequality, assuming $n\ge1$. I've been working on this one using log properties and trying to reduce this inequalitiy to simpler ones. Though!, is it even correct? or am I trying to prove a falsity? Thanks, I strongly suspect I'm wasting my time on this one??

• Let $n\rightarrow\infty$, then the left goes to $1$ and the right goes to $0$, so the inequality is wrong for large enough $n$. – J.R. Feb 25 '14 at 0:26
• It is not true, as the left-hand side tends to 1, as $n\to\infty$, while the right-hand side tends to 0. – Yiorgos S. Smyrlis Feb 25 '14 at 0:27

Notice that $n\log\left(1+\frac{1}{n}\right) - \log\left(1 + \frac{1}{n+1}\right) < \log\left(1 + \frac{1}{n+1}\right)$ can be rewritten as $n\log\left(1+\frac{1}{n}\right) - 2\log\left(1 + \frac{1}{n+1}\right) < 0$. Since $1+\frac{1}{n+1} < 1+\frac{1}{n}$ and $\log$ is an increasing function, we have that
$$n\log\left(1+\frac{1}{n}\right) - 2\log\left(1+\frac{1}{n+1}\right) > (n-2)\log\left(1+\frac{1}{n+1}\right).$$
But, for $n\ge2$, the right hand side is greater than zero, so $$n\log\left(1+\frac{1}{n}\right) - 2\log\left(1+\frac{1}{n+1}\right) > 0\\ n\log\left(1+\frac{1}{n}\right) - \log\left(1 + \frac{1}{n+1}\right) > \log\left(1 + \frac{1}{n+1}\right).$$