<s>[INST] 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 $\ln(2(m+1)) + 2 -\ln(2)$.
 [/INST] 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.
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