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

A non-induction version, just for diversification. Let's note $$S_n=\sum_{k=1}^{n} \frac{1}{k}$$
which is $$S_n=\ln(n)+\gamma +\varepsilon_n$$
Now
$$\sum_{k=1}^{n}\left ( \frac{1}{k} + \frac{2}{k+n} \right )= \sum_{k=1}^{n} \frac{1}{k} + \sum_{k=1}^{n} \frac{2}{k+n} $$
$$ = 1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n} + 2\left ( \frac{1}{n+1} + \frac{1}{n+2} + ... + \frac{1}{2n} \right ) $$
$$=S_n+2S_{2n}-2S_n=\ln(n)+\gamma +\varepsilon_n + 2\ln(2n)+2\gamma +2\varepsilon_{2n} - 2\ln(n)-2\gamma -2\varepsilon_n$$
$$=\ln(n)+\gamma + 2\ln2 + 2\varepsilon_{2n} -\varepsilon_n < $$
$$<\ln(n)+1.96352 + 2\varepsilon_{2n} -\varepsilon_n$$
And $$2\varepsilon_{2n} -\varepsilon_n \sim \frac{2}{4n} - \frac{1}{2n}$$
i.e. first $n=1..4$ can be verified "manually".
