bobbie71G

2020-11-29

Determine the convergence or divergence of the series.
$\sum _{n=1}^{\mathrm{\infty }}\left(\frac{1}{{n}^{2}}-\frac{1}{{n}^{3}}\right)$

faldduE

P-series test:
$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{n}^{p}}$
If $p>1$ series is convergent
If $p\le 1$ series is divergent
Series sum or difference rule:
If $\sum _{n=1}^{\mathrm{\infty }}{a}_{n}$ and $\sum _{n=1}^{\mathrm{\infty }}{b}_{n}$ are both convergent then
$\sum _{n=1}^{\mathrm{\infty }}\left({a}_{n}±{b}_{n}\right)=\sum _{n=1}^{\mathrm{\infty }}{a}_{n}±\sum _{n=1}^{\mathrm{\infty }}{b}_{n}$
And sum of two convergent series is also convergent
Given that $\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{n}^{p}}$
Let $\sum _{n=1}^{\mathrm{\infty }}{a}_{n}=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{n}^{2}}$ and $\sum _{n=1}^{\mathrm{\infty }}{b}_{n}=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{n}^{3}}$
Using p-series test both $\sum {a}_{n}$ and $\sum {b}_{n}$ are convergent so using series sum or difference rule $\sum _{n=1}^{\mathrm{\infty }}\left(\frac{1}{{n}^{2}}-\frac{1}{{n}^{3}}\right)=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{n}^{2}}-\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{n}^{3}}$
Since $\sum _{n=1}^{\mathrm{\infty }}{a}_{n}=\frac{1}{{n}^{2}}$ and $\sum _{n=1}^{\mathrm{\infty }}{b}_{n}=\frac{1}{{n}^{3}}$ both are convergent. Hence $\sum _{n=1}^{\mathrm{\infty }}\left(\frac{1}{{n}^{2}}-\frac{1}{{n}^{3}}\right)$ is also convergent

Jeffrey Jordon