How to prove $n \ge 2 {\left ( \sum_{k=1}^n \frac{1}{k} \right)}^2$ for all $n \ge 2$? I'm trying to prove that $$n \ge 2 {\left ( \sum_{k=1}^n \frac{1}{k} \right)}^2$$ for all $n \ge 2$. I try induction on $n$ as follows:

My attempt:
The inequality holds for $n=2$. Let it holds for $n$. Our goal is to show that $$2 {\left ( \sum_{k=1}^n \frac{1}{k} \right)}^2+1 \ge 2 {\left ( \sum_{k=1}^{n+1} \frac{1}{k} \right)}^2$$
This is equivalent to $$1 \ge \frac{1}{n+1} \left ( \frac{1}{n+1}+ \sum_{k=1}^n \frac{1}{k} \right) =\frac{1}{(n+1)^2}  + \frac{1}{n+1} \sum_{k=1}^n \frac{1}{k}$$
I'm unable to approximate the sum $\sum_{k=1}^n \frac{1}{k}$. Could you please shed me some light on the last step?
 A: Note that the statement itself tells you how to approximate $\sum_{k=1}^{n} \frac{1}{k}$, so with wishful thinking we hope it is sufficient (which need not be the case). 

Change the statement to showing that $\sqrt{\frac{n}{2}} \geq \sum_{k=1}^{n} \frac{1}{k}$.   
Then, the induction step requires us to show that $\sum_{k=1}^{n+1} \frac{1}{k} \leq \sqrt{\frac{n}{2}} + \frac{1}{n+1} \leq \sqrt{ \frac{n+1}{2}}$
Since $\sqrt{ \frac{n+1}{2}} - \sqrt{ \frac{n}{2}} = \frac{\frac{1}{2}}{\sqrt{ \frac{n+1}{2}} + \sqrt{ \frac{n}{2}}} \geq \frac{1}{2*\sqrt{2(n+1)}} \geq \frac{1}{n+1}$ when $n+1 \geq 8$.
So, start the induction at $n= 7$, and check the initial base cases. 
A: We can use that bound for harmonic series
$$\sum_{k=1}^n\frac{1}{k} \leq \ln n + 1$$
therefore
$$\frac{1}{(n+1)^2}  + \frac{1}{n+1} \sum_{k=1}^n \frac{1}{k}\le \frac{1}{(n+1)^2}  + \frac{\ln n + 1}{n+1} \le 1$$
indeed
$$ \frac{1}{(n+1)^2}  + \frac{\ln n + 1}{n+1} \le 1 \iff \frac{1}{n+1}  + \ln n + 1 \le n+1 \iff \ln n  \le n-\frac{1}{n+1}$$
which is true indeed
$$   \ln (1+(n-1))  \le n-1\le n-\frac{1}{n+1}$$
