Jamarion Kerr

2022-01-22

Need to prove the sequence:
${a}_{n}=1+\frac{1}{{2}^{2}}+\frac{1}{{3}^{2}}+\dots +\frac{1}{{n}^{2}}$ converges.

Gordon Stephens

Expert

Notice that $2{k}^{2}\ge k\left(k+1\right)⇒\frac{1}{{k}^{2}}\le \frac{2}{k\left(k+1\right)}$
$\sum _{k=1}^{\mathrm{\infty }}\frac{2}{k\left(k+1\right)}=\frac{2}{1×2}+\frac{2}{2×3}+\frac{2}{3×4}+\dots$
$\sum _{k=1}^{\mathrm{\infty }}\frac{2}{k\left(k+1\right)}=2\left(\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\dots \right)$
$\sum _{k=1}^{\mathrm{\infty }}\frac{2}{k\left(k+1\right)}=2\left(1\right)=2$
Therefore $\sum _{k=1}^{\mathrm{\infty }}\frac{1}{{k}^{2}}\le 2$

Micheal Hensley

Expert

${a}_{n}=\frac{1}{{n}^{2}}$ and because ${a}_{n}>0$ we have $|{a}_{n}|={a}_{n}$
First: check the necessary condition
$\underset{n\to \mathrm{\infty }}{lim}n{a}_{n}=\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n=0}$
Second: check DAlemberts ratio test
$\underset{n\to +\mathrm{\infty }}{lim}|\frac{{a}_{n+1}}{{a}_{n}}|=\underset{n\to +\mathrm{\infty }}{lim}\frac{{a}_{n+1}}{{a}_{n}}=\underset{n\to +\mathrm{\infty }}{lim}{\left(\frac{n}{n+1}\right)}^{2}=1$
Third: Because the answer of DAlemberts test is 1, you should use Raabes

RizerMix

Expert

This here should work with $n\ge 1$ ${s}_{n}=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{n}^{2}}=\frac{1}{1}+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{{n}^{2}}$ ${b}_{n}=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{2}^{n-1}}=\frac{1}{1}+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{{2}^{n-1}}$ ${b}_{n}$ is directly compared greater than ${s}_{n}$: ${s}_{n}<{b}_{n}$ and ${b}_{n}$ converges, because of its ratio test : $\frac{1}{{2}^{n-1+1}}/\frac{1}{{2}^{n-1}}=\frac{{2}^{n-1}}{{2}^{n}}=\frac{1}{2}<1$