Prove that $\displaystyle \sum_{k=1}^n \bigg(\dfrac{1}{k}+\dfrac{2}{k+n}\bigg ) \leq \ln(2n) + 2 -\ln(2)$ Prove that $$\displaystyle \sum_{k=1}^n \bigg(\dfrac{1}{k}+\dfrac{2}{k+n}\bigg ) \leq \ln(2n) + 2 -\ln(2).$$


I was thinking of using mathematical induction for this. That is,

We prove by induction on $n$. The case $n=1$ holds trivially since $2 \leq 2$. Now assume the result holds for some $m$. Then by assumption we know that $$\displaystyle \sum_{k=1}^{m+1} \bigg(\dfrac{1}{k}+\dfrac{2}{k+m}\bigg ) \leq \ln(2m) + 2 -\ln(2)+\dfrac{1}{m+1}+\dfrac{2}{2m+1}. $$ 

We must relate this somehow to $\ln(2(m+1)) + 2 -\ln(2)$.
First note, that we can rewrite the RHS of your inequality as $\ln(n) +2$.

We proceed by induction. We compute

$$ \sum_{k=1}^{n+1} \left( \frac{1}{k} + \frac{2}{k+n+1}\right) 
= \frac{1}{n+1} + \frac{2}{2n+2} + \sum_{k=1}^n \frac{1}{k} + \sum_{k=1}^{n} \frac{2}{k+n+1}$$

shifting the index yields

$$ =\frac{2}{n+1} + \sum_{k=1}^n \frac{1}{k} + \sum_{k=2}^{n+1} \frac{2}{k+n}
= \sum_{k=1}^n \frac{1}{k} + \sum_{k=1}^{n+1} \frac{2}{k+n}
= \sum_{k=1}^n\left( \frac{1}{k} + \frac{2}{k+n}\right) + \frac{2}{2n+1}$$

applying the induction hypothesis gives

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

Hence, we are left to show that

$$ \frac{2}{2n+1} - \ln\left(1 + \frac{1}{n}\right) = \frac{2}{2n+1} +  \ln(n) - \ln(n+1) \leq 0. $$

This is already done in a previous answer.
If you want to do a proof by induction.
you have covered the base case.

Inductive hypothesis:

$\sum_\limits{k=1}^n \bigg(\dfrac{1}{k}+\dfrac{2}{k+n}\bigg ) \leq \ln(2n) + 2 -\ln(2)$.

We need to show that:

$\sum_\limits{k=1}^{n+1} \bigg(\dfrac{1}{k}+\dfrac{2}{k+n+1}\bigg ) \leq \ln(2n) + 2 -\ln(2)$

thoughts:

$\ln(2n) + 2 -\ln(2) = \ln (n) + 2$

$\sum_\limits{k=1}^{n+1} \bigg(\dfrac{1}{k}+\dfrac{2}{k+n+1}\bigg ) = \sum_\limits{k=1}^{n} \bigg(\dfrac{1}{k}+\dfrac{2}{k+n}\bigg ) + \dfrac{2}{2n+1}$ 

$\sum_\limits{k=1}^{n} \bigg(\dfrac{1}{k}+\dfrac{2}{k+n}\bigg )\leq  \ln n + 2$ by the inductive hypothesis.

So, if $\dfrac{2}{2n+1}-\ln (n) + 2 \leq \ln(n+1) + 2$, we are done.

$\dfrac{2}{2n+1} &lt; \ln(1+\frac{1}{n})$

$\sum_\limits{k=1}^{n+1} \bigg(\dfrac{1}{k}+\dfrac{2}{k+n+1}\bigg ) \leq \ln(2n) + 2 -\ln(2)$

QED