0
$\begingroup$

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??

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

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 $$n\log\left(1+\frac{1}{n}\right) - 2\log\left(1+\frac{1}{n+1}\right) > 0\\ n\log\left(1+\frac{1}{n}\right) - \log\left(1 + \frac{1}{n+1}\right) > \log\left(1 + \frac{1}{n+1}\right).$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.