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Q. 51

Expert-verified
Found in: Page 625

### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861

# In Exercises 48–51 find all values of p so that the series converges.$\sum _{k=1}^{\infty }\frac{k}{{\left(1+{k}^{2}\right)}^{p}}$

$\mathrm{The}\mathrm{integral}{\int }_{\mathrm{x}=1}^{\infty }\frac{\mathrm{x}}{{\left(1+{\mathrm{x}}^{2}\right)}^{\mathrm{p}}}\mathrm{dx}\mathrm{converges}\mathrm{for}\mathrm{p}>1.\phantom{\rule{0ex}{0ex}}\mathrm{Thus},\mathrm{the}\mathrm{series}\sum _{\mathrm{k}=1}^{\infty }\frac{\mathrm{k}}{{\left(1+{\mathrm{k}}^{2}\right)}^{\mathrm{p}}}\mathrm{is}\mathrm{convergent}\mathrm{for}\mathrm{p}>1.$

See the step by step solution

## Step 1. Given information is:

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

## Step 2. Examining nature of given function:

$\mathrm{Consider}\mathrm{the}\mathrm{function}\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{x}}{{\left(1+{\mathrm{x}}^{2}\right)}^{\mathrm{p}}}.\phantom{\rule{0ex}{0ex}}\mathrm{The}\mathrm{function}\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{x}}{{\left(1+{\mathrm{x}}^{2}\right)}^{\mathrm{p}}}\mathrm{is}\mathrm{continuous},\mathrm{decreasing},\mathrm{with}\mathrm{positive}\mathrm{terms}.\phantom{\rule{0ex}{0ex}}\mathrm{Therefore}\mathrm{all}\mathrm{the}\mathrm{conditions}\mathrm{of}\mathrm{integral}\mathrm{test}\mathrm{are}\mathrm{fulfilled}.\phantom{\rule{0ex}{0ex}}\mathrm{So},\mathrm{integral}\mathrm{test}\mathrm{is}\mathrm{applicable}.$

## Step 3. Solving the integral:

$\mathrm{Consider}\mathrm{the}\mathrm{integral}:{\int }_{\mathrm{x}=1}^{\infty }\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}={\int }_{\mathrm{x}=1}^{\infty }\frac{\mathrm{x}}{{\left(1+{\mathrm{x}}^{2}\right)}^{\mathrm{p}}}\mathrm{dx}.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Therefore},\phantom{\rule{0ex}{0ex}}{\int }_{\mathrm{x}=1}^{\infty }\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\underset{\mathrm{k}\to \infty }{\mathrm{lim}}{\int }_{\mathrm{x}=1}^{\mathrm{k}}\frac{\mathrm{x}}{{\left(1+{\mathrm{x}}^{2}\right)}^{\mathrm{p}}}\mathrm{dx}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\underset{\mathrm{k}\to \infty }{\mathrm{lim}}{\int }_{\mathrm{u}=2}^{{\mathrm{k}}^{2}+1}\frac{\mathrm{du}}{{\mathrm{u}}^{\mathrm{p}}}\left(\mathrm{Put}1+{\mathrm{x}}^{2}=\mathrm{u},⇒2\mathrm{xdx}=\mathrm{du}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\underset{\mathrm{k}\to \infty }{\mathrm{lim}}{\left[\frac{{\mathrm{u}}^{-\mathrm{p}+1}}{-\mathrm{p}+1}\right]}_{2}^{{\mathrm{k}}^{2}+1}\left(\mathrm{Integrating}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{1}{2\left(-\mathrm{p}+1\right)}\underset{\mathrm{k}\to \infty }{\mathrm{lim}}{\left[\frac{1}{{\mathrm{u}}^{\mathrm{p}-1}}\right]}_{2}^{{\mathrm{k}}^{2}+1}\left(\mathrm{Substitution}\right)\phantom{\rule{0ex}{0ex}}$

## Step 4. Result:

$\mathrm{The}\mathrm{improper}\mathrm{integral}\mathrm{converges}\mathrm{to}\mathrm{finite}\mathrm{value}\mathrm{only}\mathrm{when}\mathrm{p}>1.\phantom{\rule{0ex}{0ex}}\mathrm{Therefore},\mathrm{the}\mathrm{integral}{\int }_{\mathrm{x}=1}^{\infty }\frac{x}{{\left(1+{x}^{2}\right)}^{p}}\mathrm{dx}\mathrm{converges}\mathrm{for}\mathrm{p}>1.\phantom{\rule{0ex}{0ex}}\mathrm{Thus},\mathrm{the}\mathrm{series}\sum _{\mathrm{k}=1}^{\infty }\frac{k}{{\left(1+{k}^{2}\right)}^{p}}\mathrm{is}\mathrm{convergent}\mathrm{for}\mathrm{p}>1.$