If $n$ is a positive integer, Prove that $\frac1{2^2}+\frac1{3^2}+\dotsb+\frac1{n^2}\lt\frac{2329}{3600}.$

Question

If $n$ is a positive integer, Prove that

$$\frac1{2^2}+\frac1{3^2}+\dotsb+\frac1{n^2}\lt\frac{2329}{3600}.$$

please don't refer to the famous $1+\frac1{2^2}+\frac1{3^2}+\dotsb=\frac{\pi^2}6$.

I am looking a method that doesn't use $\text{“}\pi\text{''}.$

Unfortunately, I know and tried only $\text{“}\pi\text{''}$ method.

Answer 1

I like this kind of numerical challenges. Let us exploit creative telescoping.
For any $n\geq 2$ we have $$ \frac{1}{n^2}-\left[\frac{1}{n}-\frac{1}{n+1}\right]=\frac{1}{n^2(n+1)}\tag{1} $$ $$ \frac{1}{n^2(n+1)}-\left[\frac{1}{2n^2}-\frac{1}{2(n+1)^2}\right]=\frac{1}{2n^2(n+1)^2}\tag{2} $$ $$ \frac{1}{2n^2(n+1)^2}-\left[\frac{1}{6n^3}-\frac{1}{6(n+1)^3}\right]=-\frac{1}{6n^3(n+1)^3}\tag{3} $$ $$ \frac{1}{6n^3(n+1)^3}-\left[\frac{1}{30n^5}-\frac{1}{30(n+1)^5}\right]=\frac{5n^2+5n+1}{30n^5(n+1)^5}\tag{4}$$ $$ \frac{5n^2+5n+1}{30n^5(n+1)^5}< \frac{1}{42 n^7}-\frac{1}{42(n+1)^7}\tag{5} $$ hence by setting $g(n)=\frac{1}{n}+\frac{1}{2n^2}+\frac{1}{6n^3}-\frac{1}{30n^5}$ we have $$\begin{eqnarray*} \zeta(2)-1=\sum_{n\geq 2}\frac{1}{n^2} &=& \sum_{n\geq 2}\left[g(n)-g(n+1)\right]+\sum_{n\geq 2}\frac{5n^2+5n+1}{30n^5(n+1)^5}\\ &<& g(2)+\frac{1}{42\cdot 2^7}=\color{red}{\frac{5779}{8960}}<\frac{2322}{3600}.\tag{6}\end{eqnarray*} $$

The same technique proves Stirling's inequality and the "acceleration formula" $$ \sum_{n\geq 1}\frac{1}{n^2} = \sum_{n\geq 1}\frac{3}{n^2\binom{2n}{n}}\tag{7} $$ leading to the more accurate approximation $\zeta(2)-1\approx \frac{51077}{79200}$.

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There also is a simple and accurate lower bound. By choosing $h(x)=\frac{6x-3}{6x^2-6x+2}$ we have $\frac{1}{x^2}\geq h(x)-h(x+1)$ for any $x\geq 2$, with the difference behaving like $\frac{1}{9x^6}$. It follows that $$ \zeta(2)-1\geq \color{red}{\frac{9}{14}} = h(2).\tag{8}$$ By selecting $k(x)=\frac{60x^2-60x+31}{60x^3-90x^2+66x-18}$ we get $\zeta(2)-1\leq \frac{151}{234}$, simpler but less accurate than $(6)$.

Answer 2

A viable approach is also to apply a "delayed" creative telescoping based on the identity $\frac{1}{n^2}<\frac{1}{n^2-1}=\frac{1}{2}\left(\frac{1}{n-1}-\frac{1}{n+1}\right)$. This approach leads to

$$ \zeta(2)-1<\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}\right)+\frac{1}{2}\sum_{n>5}\left(\frac{1}{n-1}-\frac{1}{n+1}\right)=\frac{2329}{3600} $$ in a single line.

Answer 3

Let $\displaystyle S = \sum_{n=2}^\infty \frac 1 {n^2}.\,$ A simple integral test shows $S<\infty.$

Let $T$ be some number a little bit bigger than $S.$ Then \begin{align} & \frac 1 {2^2} + \cdots + \frac 1 {n^2} \\[10pt] < {} & \frac 1 {2^2} + \cdots + \frac 1 {n^2} + \int_n^\infty \frac{dx}{x^2} \\[10pt] = {} & \left(\frac 1 {2^2} + \cdots + \frac 1 {n^2} \right) + \frac 1 n \tag 1 \\[10pt] < {} & S + \frac 1 n \\[10pt] < {} & T \text{ if $n$ is big enough.} \end{align} This shows $\displaystyle \frac 1 {2^2} + \cdots + \frac 1 {n^2} < T$ if $n$ is big enough, but that sum is smaller if $n$ is smaller; therefore that inequality holds without any qualification on the size of $n.$

Thus we only need to know how big $T$ needs to be, or how small $T$ can be. I think maybe if you take $n=500$ then line $(1)$ might show that $T=2329/3600$ can serve.

This is computation-intensive. Maybe there's also an intelligent way to do it.

Answer 4

For any natural $n$, we have

$$S_1=\sum_{k=2}^\infty\frac1{k^2}>\sum_{k=2}^n\frac1{k^2}$$

Likewise, consider the following alternating sum:

$$S_2=\sum_{k=2}^\infty\frac{(-1)^k}{k^2}$$

And see that we have

$$S_1+S_2=\sum_{k=2}^\infty\frac{1+(-1)^k}{k^2}=\sum_{k=1}^\infty\frac2{(2k)^2}=\frac12+\frac12S_1$$

Thus, we have

$$S_1=1-2S_2$$

Furthermore, it is easy to see that

$$S_2>\sum_{k=2}^{23}\frac{(-1)^k}{k^2}$$

And thus,

$$S_1<1-2\sum_{k=2}^{23}\frac{(-1)^k}{k^2}<\frac{2329}{3600}$$

Not a terrible sum to evaluate there.

Answer 5

approximate by telescoping series. $\left(\sum_{k=2}^{20} \dfrac1{k^2}\right)+(\frac1{20}-\frac1{21})+(\frac1{21}-\frac1{22})+・・・(\frac1{n-1}-\frac1n)=.62976+\frac1{20}-\frac1n=.64616-\frac1n<0.6469\dot4=\frac{2329}{3600}$.

Also $16$ instead of 20 holds.

Answer 6

The simplest approach is to approximate the sum by the integral. The crude way of doing it just uses the fact that $1/x^2$ is a decreasing function, so the integral from $k$ to $k+1$ is bigger than $1/(k+1)^2.$ to make it sharper, one notes that since $1/x^2$ is convex, the integral is actually bigger than $\frac12(1/n^2 + 1/(n+1)^2).$