An inequality with $\sum_{k=2}^{n}\left(\frac{2}{k}+\frac{H_{k}-\frac{2}{k}}{2^{k-1}}\right)$

Question

show that $$\sum_{k=2}^{n}\left(\dfrac{2}{k}+\dfrac{H_{k}-\frac{2}{k}}{2^{k-1}}\right)\le 1+2\ln{n}$$where $ n\ge 2,H_{k}=1+\dfrac{1}{2}+\cdots+\dfrac{1}{k}$

Maybe this $\ln{k}<H_{k}<1+\ln{k}$?

Answer 1

Hint:Note $$\dfrac{2}{k}=2H_{k}-2H_{k-1}$$ then $$\sum_{k=2}^{n}\dfrac{H_{k}-\frac{2}{k}}{2^{k-1}} =\sum_{k=2}^{n}\dfrac{2H_{k-1}-H_{k}}{2^{k-1}} =\sum_{k=2}^{n}\left(\dfrac{H_{k-1}}{2^{k-2}}-\dfrac{H_{k}}{2^{k-1}}\right)$$

Answer 2

We have, by partial summation: $$\begin{eqnarray*} \sum_{k=2}^{n}\frac{H_k}{2^{k-1}}&=&H_n\left(1-\frac{1}{2^{n-1}}\right)-\sum_{k=2}^{n-1}\left(1-\frac{1}{2^{k-1}}\right)\frac{1}{k+1}\\&=&1-\frac{H_n}{2^{n-1}}+\sum_{k=2}^{n}\frac{2}{k\, 2^{k-1}}\tag{1}\end{eqnarray*}$$ hence: $$ \sum_{k=2}^{n}\frac{H_k-\frac{2}{k}}{2^{k-1}}=1-\frac{H_n}{2^{n-1}}\tag{2}$$ and: $$\sum_{k=2}^{n}\left(\frac{2}{k}+\frac{H_k-\frac{2}{k}}{2^{k-1}}\right)\leq 2 H_n-1\leq 2\log n+1.\tag{3}$$