How can I show the following?
$$\sum_{n=1}^\infty \left(\frac{1}{n}\right)^2<\infty$$
Isn't it: $$\sum_{n=1}^\infty \left(\frac{1}{n}\right)^2=\left(\frac{1}{1}\right)^2+\left(\frac{1}{2}\right)^2+\cdots+\left(\frac{1}{\infty}\right)^2=\infty$$
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3See Bassel problem – Feb 12 '20 at 16:07
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2The term $1/\infty$ is wrong, because it never appears in the sum. – Dietrich Burde Feb 12 '20 at 16:09
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1Note: $\sum\limits_{n=1}^\infty2^{-n}\ne\infty$ too – J. W. Tanner Feb 12 '20 at 16:09
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1On the note of J.W.Tanner: You can easily see this by taking a piece of paper, splitting it in half, splitting one half in half again and so on. You should even be able to guess what the limit will be :) – Luke Feb 12 '20 at 16:11
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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.
cansomeonehelpmeout
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$$S_N=\sum_{k=1}^N\dfrac{1}{k^2}=\dfrac{\pi^2}{6}-\psi'(N+1)$$ where $\psi'$ is the first derivative of the digamma function. Because $$\lim_{N\to+\infty}\psi'(N+1)=0,$$ $S_N \to \dfrac{\pi^2}{6}$.
Gary
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Riccardo.Alestra
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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.
Ross Millikan
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