How prove this inequality $\sum_{k=2}^{49}\frac{1}{k^2}\ge\frac{9}{10}\ln{2}$? show that
$$\sum_{k=2}^{49}\dfrac{1}{k^2}\ge\dfrac{9}{10}\ln{2}$$
my idea: since
$$\dfrac{1}{k^2}\ge\dfrac{1}{k(k+1)}=\dfrac{1}{k}-\dfrac{1}{k+1}$$
so
$$\sum_{k=2}^{49}\dfrac{1}{k^2}\ge\sum_{k=2}^{49}\left(\dfrac{1}{k}-\dfrac{1}{k+1}\right)=\dfrac{1}{2}-\dfrac{1}{50}=\dfrac{12}{25}$$ 
But $$\dfrac{9}{10}\ln{2}>\dfrac{12}{25}$$
other idea:
$$\dfrac{1}{k^2}\ge\dfrac{3}{(2k-1)(2k+1)}$$
so How prove it? Thank you
and I found the constant $\dfrac{9}{10}\ln{2}$ is best value.can see
http://www.wolframalpha.com/input/?i=%28%5Csum_%7Bk%3D2%7D%5E%7B49%7D%5Cdfrac%7B1%7D%7Bk%5E2%7D%29-%289%2F10%29*ln2&dataset=

I Looking forward to answer this problem by hand! Thank you

$$\sum_{k=2}^{49} \dfrac{1}{k^2} = \sum_{k=1}^\infty \dfrac{1}{k^2} - 1 - \sum_{k=50}^\infty \dfrac{1}{k^2} = \dfrac{\pi^2}{6} - 1 - \sum_{k=50}^\infty \dfrac{1}{k^2}$$.
The function $f(x) = 1/x^2$ is strictly decreasing on $[49, +\infty)$ so the right hand Riemann sum $\sum_{k=50}^\infty \dfrac{1}{k^2}$ is less than $\int_{49}^\infty \dfrac{1}{x^2}\, dx$. The latter is equal to $1/49$. Therefore,
$$\dfrac{\pi^2}{6} - 1 - \sum_{k=50}^\infty \dfrac{1}{k^2} > \dfrac{\pi^2}{6} - 1 - \dfrac{1}{49}$$
which one can check is numerically bigger than $\dfrac{9}{10} \log(2)$ by about $0.00069344$. This is not a direct estimate but perhaps of some help.
