Show that for all integers $n\geq 2$ that $\sum_{k=1}^n \frac{1}{k^2} \geq \frac{3n}{2n+1}$

Question

It can be done with induction by comparing $\frac{3n}{2n+1}+\frac{1}{(n+1)^2}$ and $\frac{3(n+1)}{2(n+1)+1}$, but I would like a better alternative since this one involves a lot of expanding and is just algebra, and it's not very impressive in my opinion.

Answer 1

Assuming that you know about generalized harmonic numbers $$\sum_{k=1}^n \frac{1}{k^2}=H_n^{(2)}$$ and its expansion is $$H_n^{(2)}=\frac{\pi ^2}{6}-\frac{1}{n}+\frac{1}{2 n^2}+O\left(\frac{1}{n^3}\right)$$ while $$\frac{3n}{2n+1}=\frac{3}{2}-\frac{3}{4 n}+\frac{3}{8 n^2}+O\left(\frac{1}{n^3}\right)$$ So $$\text{lhs - rhs}=\left(\frac{\pi ^2}{6}-\frac{3}{2}\right)-\frac{1}{4 n}+\frac{1}{8 n^2}+O\left(\frac{1}{n^3}\right)$$ which is always positive.

Answer 2

First, I would begin plugging in values starting from 2 and going up. What we notice immediately is that $\sum\limits_{k=1}^n(\frac{1}{k^2})=1+\frac{1}{4}+\frac{1}{9}...\frac{1}{k^2}$ which would ultimately have a limit of $\frac{\pi^2}{6}$ as $k\to\infty$.

To prove that $\sum\limits_{k=1}^n(\frac{1}{k^2})\geq\frac{3n}{2n+1}$, simply need to prove that this fraction never exceeds $\frac{\pi^2}{6}$. Set up the expression and set it equal to $\frac{\pi^2}{6}$ like this:

$$ \frac{\pi^2}{6}=\frac{3n}{2n+1} $$

Through basic algebra, we get that $2n(\pi^2-9)-\pi=0$ which, through some more manipulation, shows that $n=\frac{-\pi^2}{2(\pi^2-9)}\approx-5.67...$. Because this is a negative number and the only solution to setting it equal to $\frac{\pi^2}{6}$, we know that $n\geq2$ will never exceed $\frac{\pi^2}{6}$ & ergo never be greater than $\sum\limits_{k=1}^n(\frac{1}{k^2})$.

QED qve $\sum\limits_{k=1}^n(\frac{1}{k^2})\geq\frac{3n}{2n+1}$