Why is the sum infinite fractions not infinite?

Question

How can I show the following?

$$\sum_{n=1}^\infty \left(\frac{1}{n}\right)^2<\infty$$

Isn't it: $$\sum_{n=1}^\infty \left(\frac{1}{n}\right)^2=\left(\frac{1}{1}\right)^2+\left(\frac{1}{2}\right)^2+\cdots+\left(\frac{1}{\infty}\right)^2=\infty$$

Answer 1

You could make the same argument with $$S=\sum_{i=}^\infty\frac{1}{2^i}=\frac{1}{2}+\frac{1}{4}+\dots+\frac{1}{2^\infty}$$ But it is clear from the picture below that $S=1$.

Also, in your case it is increasing and bounded by $$\sum_{k=1}^n \frac1{k ^ 2} \lt 1 + \sum_{k=2}^n \frac 1 {k (k - 1)} = 2 - \frac 1 n \lt 2$$ and therefore converges.

Answer 2

If the terms of the sum decrease quickly enough the sum will converge to a finite limit. An easier one to sum is $$\sum_{i=1}^\infty \frac 1{2^i}$$ The definition of an infinite sum is to make the upper limit $n$, then take the limit as $n \to \infty$. If the limit converges, that is the sum of the series. In this case $$\lim_{n \to \infty}\sum_{i=1}^n \frac 1{2^i}=\lim_{n \to \infty}\left(1-\frac 1{2^n}\right)=1$$ In your expression the sum also converges, but proving the limit is harder.

Answer 3

$$S_N=\sum_{k=1}^N\dfrac{1}{k^2}=\dfrac{\pi^2}{6}-\psi'(N+1)$$ where $\psi'$ is the first derivative of the digamma function. Because $$\lim_{N\to+\infty}\psi'(N+1)=0,$$ $S_N \to \dfrac{\pi^2}{6}$.