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Is there a way to sum up the series with nth term $$ x_n=1/n^2$$ I am high school student and I tried my level best to find a method to sum it up but failed.If there is a way to find this sum can this be generalised to find the sum of other series.

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This is not a proof. But this demonstrates how Euler thought about it.

$$1+\frac14+\frac19+\frac1{16}+\frac1{25}+\cdots=\frac{\pi^2}{6}$$ Euler discovered the above identity while thinking about polynomials. Bear in mind that any polynomial can be express as either a sum or a product: $$2x^3+3x^2-11x-6=2(x-2)(x+3)(x+1/2).$$ Given the form on the left, you need only find the zeros of the polynomial to obtain the form on the right. Euler wondered whether this might work with the following infinite polynomial: $$\sin x=x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}+\cdots$$ The zeros of this function are just $0,\pm\pi,\pm 2\pi,\cdots.$ so (Euler argue that) perhaps there is a product for sine of the form $$\sin x=Ax(x-\pi)(x+\pi)(x-2\pi)(x+2\pi)\cdots$$ The tricky part is finding the constant $A$. usually $A$ is the coefficient of the highest-order term, but the power series for sine has no highest-order term. Indeed, since the coefficients of $x^n$ are approaching zero, there is a good argument that $A$ should be infinitesimal. After a bit of thought, Euler spotted a solution to this problem. We know that: $$\lim_{x\to 0}\dfrac{\sin x}{x}=1.$$ But clearly: $\lim_{x\to 0}\dfrac{x(x-\pi)(x+\pi)(x-2\pi)(x+2\pi)\cdots}{x}=(-\pi)(\pi)(-2\pi)(2\pi)\cdots$ So $A$ should be the reciprocal of the infinite product of the right. This leads to the following guess: $$\sin x=x\left(1-\frac x{\pi}\right)\left(1+\frac x{\pi}\right)\left(1-\frac x{2\pi}\right)\left(1+\frac x{2\pi}\right)⋯=x\prod_{n=1}^{\infty}\left(1-\dfrac{x^2}{n^2\pi^2}\right)$$ The amazing thing is that this infinite product formula actually works!
It’s called the Euler-Wallis formula for sine, Wallis presumably being the first to prove it rigorously. Today it is known as a corollary of the Weierstrass factorization theorem. Euler, however, was more interested in the relationship between the infinite product and the Taylor series. Consider the equation: $$x\left(1-\frac{x^2}{\pi^2}\right)\left(1-\frac{x^2}{4\pi^2}\right)\left(\frac{x^2}{9\pi^2}\right)\cdots=x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}+\cdots$$ Presumably the infinite product on the right should multiply out to give the infinite sum on the left. Equating the coefficients of $x^3$ gives the following remarkable identity: $$\sum_{n=1}^{\infty}\dfrac{1}{n^2}=\dfrac{\pi^2}{6}.$$

(PS: I found this argument from someone's blog. But I forget the exact person. Sorry for that.)

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    $\begingroup$ I enjoyed reading this as it hints at the motivation behind the series. Thank you. $\endgroup$ – Karl Mar 8 '16 at 20:04
  • $\begingroup$ See this. You will find more proofs for this fact. $\endgroup$ – Bumblebee Mar 11 '16 at 10:39
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I'm sure this has already been said. But what you have there is known as the Basel Problem.

This is basically the sum:

$${1}+ {1\over 2^2} + {1\over 3^2} + {1\over 4^2}+...$$

It can also be written as:

$$\sum_{n=1}^{\infty} {1\over n^2}$$

This is known as the Zeta Function evaluated at $\space 2$. Which is written using the greek letter Zeta, which is $\zeta$.

Mathematician $\space$Leonhard Euler $\space$ gave a proof for this which shows that:

$$\zeta (2)=\sum_{n=1}^{\infty} {1\over n^2}={1}+ {1\over 2^2} + {1\over 3^2} + {1\over 4^2}+...={\pi^2 \over 6}$$

Here are some links which talk about this proof:

Link 1

Link 2

Link 3 (Video)

It's great to see young minds have this curious view on mathematics rather than the common "When is Math class going to end?" attitude. This is how the greats think.

Keep it up and always question things. You never know what you may discover.

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You can also consider that the sum corresponds to generalized harmonic numbers; so $$S_n=\sum_{i=1}^n \frac 1 {i^2}=H_n^{(2)}$$ For large values of $n$, the asymptotic expansion is $$S_n=\frac{\pi ^2}{6}-\frac{1}{n}+\frac{1}{2 n^2}-\frac{1}{6 n^3}+O\left(\frac{1}{n^4}\right)$$

Let us try using $n=10$ which is not a large number. The exact summation would lead to $\frac{1968329}{1270080}\approx 1.549767731$ while the formula would give $\frac{\pi ^2}{6}-\frac{571}{6000}\approx 1.549767400$.

This can be generalized to $$\sum_{i=1}^n \frac 1 {i^k}=H_n^{(k)}$$ for which nice formulae are obtained when $k$ is even (for odd values of $k$, this would involve $\zeta(k)$ you will learn sooner or later).

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