Inequality involving sums of reciprocals and n-th root

Question

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.

Answer 1

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) $$\left(1 + \frac{1}{n}\right) \left(1 + \frac{1}{n+1}\right) \dots \left(1 + \frac{1}{2n-1} \right) = 2.$$

Answer 2

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, $$ n2^{1/n}-n\le\sum_{k=n}^{2n-1}\frac1k $$