# Inequality involving sums of reciprocals and n-th root

I'm trying to prove this inequality. Let n be a positive integer. Prove that: $$\frac{1}{n}+\frac{1}{n+1}+\dots+\frac{1}{2n-1}\ge n\sqrt[n]{2}-n$$ I've tried doing it with algebraic, geometric, and harmonic mean, the integral bound (and realised both sides are decreasing and going towards $\log 2$), but none have worked. I'd be grateful if anyone could show me the trick needed.

• Have you tried using induction? – Hawk Jan 24 '14 at 10:41
• @Hawk: have you tried using induction for this problem? Unless I am missing something obvious, this is a red herring. – robjohn Jan 24 '14 at 15:27
• @robjohn Obviously you are not missing anything...and I did not mean to use induction directly to the problem but after some manipulations because if we use induction here, then we would have to prove yet another inequality. – Hawk Jan 24 '14 at 18:14

Write this as $$2^{1/n} \le \frac{\left(1 + \frac{1}{n}\right) + \dots + \left(1 + \frac{1}{2n-1}\right)}{n}.$$ The result follows from AM-GM once you show that (for example by induction) $$\left(1 + \frac{1}{n}\right) \left(1 + \frac{1}{n+1}\right) \dots \left(1 + \frac{1}{2n-1} \right) = 2.$$
• How did you ever come up with doing AM-GM on the numbers $1+\frac1k$? I know, once you say their product is $2$ it becomes obvious, but you should already know this trick to be able to invent this, no? – punctured dusk Jan 24 '14 at 10:50
• @barto: Well, the power $1/n$ hints at AM-GM. To use AM-GM you want some approximately same size numbers. From this point of view distributing the $n$ evenly among the $n$ terms makes sense. – J. J. Jan 24 '14 at 10:52
Since $\log(1+x)$ is concave, Jensen's inequality says that \begin{align} \frac1n\log(2) &=\frac1n\sum_{k=n}^{2n-1}\log\left(1+\frac1k\right)\\ &\le\log\left(1+\frac1n\sum_{k=n}^{2n-1}\frac1k\right)\\ \end{align} which is equivalent to $$2^{1/n}\le1+\frac1n\sum_{k=n}^{2n-1}\frac1k$$ and therefore, $$n2^{1/n}-n\le\sum_{k=n}^{2n-1}\frac1k$$