Jamarion Kerr

Answered

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.

Answer & Explanation

Gordon Stephens

Expert

2022-01-23Added 10 answers

Notice that $2{k}^{2}\ge k(k+1)\Rightarrow \frac{1}{{k}^{2}}\le \frac{2}{k(k+1)}$

$\sum _{k=1}^{\mathrm{\infty}}\frac{2}{k(k+1)}=\frac{2}{1\times 2}+\frac{2}{2\times 3}+\frac{2}{3\times 4}+\dots$

$\sum _{k=1}^{\mathrm{\infty}}\frac{2}{k(k+1)}=2((1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+\dots )$

$\sum _{k=1}^{\mathrm{\infty}}\frac{2}{k(k+1)}=2\left(1\right)=2$

Therefore$\sum _{k=1}^{\mathrm{\infty}}\frac{1}{{k}^{2}}\le 2$

Therefore

Micheal Hensley

Expert

2022-01-24Added 10 answers

First: check the necessary condition

Second: check DAlemberts ratio test

Third: Because the answer of DAlemberts test is 1, you should use Raabes

RizerMix

Expert

2022-01-27Added 437 answers

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 :
$\frac{1}{{2}^{n-1+1}}/\frac{1}{{2}^{n-1}}=\frac{{2}^{n-1}}{{2}^{n}}=\frac{1}{2}<1$

Most Popular Questions