# Prove that $1.49<\sum_{k=1}^{99}\frac{1}{k^2}<1.99$

It can be proven by induction that $$\sum_{k=1}^{n}\frac{1}{k^2}\leq2-\frac{1}{n}$$

From here, we can easily acquire the upper bound of the sum $$\sum_{k=1}^{99}\frac{1}{k^2}$$ letting $n=100$.

However, I am not quite sure about the lower bound. The standard method of constructing lower rectangles of unit width on the curve $y=\frac{1}{x^2}$ yields a lower bound of $0.99$, which isn't tight enough unfortunately. What could I do to achieve a sharper bound?

I know that as $n\rightarrow \infty$, the sum converges to $\frac{\pi^2}{6}$, but I feel that proving that result, just for this inequality, is a bit much.

• One thing that might work is this: Don't apply any "method" for the first few terms - say 1 + 1/2^2 = 5/4, period. Apply whatever method to the sum of the remaining terms. The point being those approximations are probably very bad for the first few terms... – David C. Ullrich Aug 25 '15 at 18:01
• Rectangular strips... you're on the right track! Just isolate the first strip and compare the remaining 98 with the curve. See my answer below. – hypergeometric Aug 25 '15 at 18:25

This approach as inspired by the Riemann sum. Basically, comparing the area of rectangular strips of unit width and height $\frac 1{k^2}$ with the area under the curve $y=\frac 1{k^2}$.

\begin{align} 1+\int_2^{100}\frac 1{k^2}dk<&\sum_{k=1}^{99}\frac 1{k^2}<1+\int_1^{99}\frac 1{k^2}dk\\ 1.49<&\sum_{k=1}^{99}\frac 1{k^2}<1.99 \end{align}

• +1: curiously the numerical values are the same than in my (telescoping sum) answer! – Raymond Manzoni Aug 25 '15 at 18:33
• Well done! A +1 ... this embedded the "spirit" of the question I believe – Mark Viola Aug 25 '15 at 18:59
• @Dr.MV - Thank you! Very kind of you to say so :) – hypergeometric Aug 26 '15 at 6:18
• @RaymondManzoni - Thank you! Yes, I did notice the curious coincidence. Or is it a coincidence at all? – hypergeometric Aug 26 '15 at 6:19
• Well of course not since $\int_n^{n+1}\frac {dk}{k^2}=\frac 1{n\,(n+1)}$ but neat anyway. Cheers, – Raymond Manzoni Aug 26 '15 at 6:53

Hint: Telescoping and $$1+\sum_{k=2}^{99}\frac{1}{k\;(k+1)}<\sum_{k=1}^{99}\frac{1}{k^2}<1+\sum_{k=2}^{99}\frac{1}{k\;(k-1)}$$

For the lower bound, just compute the first few terms: $$\sum_{k=1}^{6}\frac{1}{k^2} \approx 1.4913 > 1.49$$

The exact value is $$\sum_{k=1}^{6}\frac{1}{k^2} = \frac{5369}{3600} = 1.4913\bar8$$

Following the creative telescoping approach I used to solve another question,

$$\frac{1}{m}+\frac{1}{2m^2}+\frac{1}{6m^3}-\frac{1}{30m^5}\leq\sum_{n\geq m}\frac{1}{n^2}\leq\frac{1}{m}+\frac{1}{2m^2}+\frac{1}{6m^3}\tag{1}$$ hence: $$\frac{1}{m}-\frac{1}{2m^2}+\frac{1}{6m^3}-\frac{1}{30m^5}\leq\sum_{n> m}\frac{1}{n^2}\leq\frac{1}{m}-\frac{1}{2m^2}+\frac{1}{6m^3}\tag{2}$$ and by setting $m=100$ we have: $$0.009950166663\ldots\leq\frac{\pi^2}{6}-H_{100}^{(2)}\leq 0.009950166666\ldots\tag{3}$$ so:

$$\sum_{n=1}^{99}\frac{1}{n^2}=\color{red}{1.6348839}+E,\qquad |E|\leq \color{red}{10^{-9}}. \tag{4}$$