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I have recently learnt to find some limits and values of some series using integration. This is called as evaluating series and limits using limit as a sum in my textbook.

Now, I tried to derive this series using what I learnt. (It is not part of my textbook. I was just doing this for my own fun):

$$\sum_{r=1}^\infty\frac{1}{r^2}= \frac{π^2}{6}$$

Here is what I did :

Let $S = \lim_{n\to\infty}\sum_{r=1}^n\frac{1}{r^2}$

Using change of variable of $n\to n^2$ and $r\to r^2$, call $n^2 = x$ and $r^2 = y$, we have :

$$S=\lim_{x\to\infty}\frac{1}{x}\sum_{y=1}^x\frac{x}{y}$$

Now using limit as a sum formula, we have :

$$S = \int_0^1\frac{1}{x}dx=\infty$$

Of course this is wrong.

Where did I went wrong? Can you provide a correct proof of this series using this limit as a sum integration method?

(There is a famous question of this series on this site but on the answers to that, I didn't find a solution involving limit as a sum method. Hence my question is different.)

An_Elephant
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2 Answers2

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In the equation:$$S=\lim_{x\to\infty}\frac{1}{x}\sum_{y=1}^x\frac{x}{y}$$The index of the sum should've been $\sqrt{y}$, which is a bit complicated to work with.

Kamal Saleh
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  • If we plug back our substitution values, we get : $\lim_{x\to\infty}\frac{1}{x}\sum_{y=1}^x\frac{x}{y}=$$\lim_{n^2\to \infty} \sum_{r^2 = 1}^n^2 \frac{1}{r^2}$ – An_Elephant Dec 19 '22 at 15:39
  • @An_Elephant Yes, you are right. This is why the index should've been $\sqrt{y}$, which technically could happen, nothing wrong about that, but it might be the reason why the sum diverges. – Kamal Saleh Dec 19 '22 at 15:41
  • I also thought that the wrong-ness would be in substitution. However, I still can't understand why is it because because variables reach same value of limit. Can you elaborate more in the answer. Thanks – An_Elephant Dec 19 '22 at 15:45
  • @An_Elephant Your substitution for $x$ is fine since $\sqrt{x}\rightarrow\infty$ so $x\rightarrow\infty$, but when you change the index of the sum from $y=1$ to $\sqrt{y}=1$ you actually get more terms. – Kamal Saleh Dec 19 '22 at 16:11
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you could verify below equation using Taylor's expansion

$$\sum_{r=1}^\infty{1\over r^2}=-\int_{0}^1{{ln(1-x)\over x}dx}$$

by substituting $$ln(1-x)=z$$

we get, $$=\int_0^\infty \frac{z}{e^z-1}dz$$

which is called BOSE INTEGRAL and it is solved using complex analysis

a useful link for solving the same problem

CaptainPotato
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