bobbie71G

2020-11-29

Determine the convergence or divergence of the series.

$\sum _{n=1}^{\mathrm{\infty}}(\frac{1}{{n}^{2}}-\frac{1}{{n}^{3}})$

faldduE

Skilled2020-11-30Added 109 answers

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

$\sum _{n=1}^{\mathrm{\infty}}({a}_{n}\pm {b}_{n})=\sum _{n=1}^{\mathrm{\infty}}{a}_{n}\pm \sum _{n=1}^{\mathrm{\infty}}{b}_{n}$

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 $\sum _{n=1}^{\mathrm{\infty}}(\frac{1}{{n}^{2}}-\frac{1}{{n}^{3}})=\sum _{n=1}^{\mathrm{\infty}}\frac{1}{{n}^{2}}-\sum _{n=1}^{\mathrm{\infty}}\frac{1}{{n}^{3}}$

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 $\sum _{n=1}^{\mathrm{\infty}}(\frac{1}{{n}^{2}}-\frac{1}{{n}^{3}})$ is also convergent

If

If

Series sum or difference rule:

If

And sum of two convergent series is also convergent

Given that

Let

Using p-series test both

Since

Jeffrey Jordon

Expert2021-12-27Added 2605 answers

Answer is given below (on video)

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