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 $\ln(2(m+1)) + 2 -\ln(2)$.

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