Q. 51 - Calculus

Question

In Exercises 48–51 find all values of p so that the series converges.

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

Step-by-Step Solution

Answer

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

Step 1. Given information is:

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

Step 2. Examining nature of given function:

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

Step 3. Solving the integral:

$$\begin{gathered} \mathrm{Consider} \mathrm{the} \mathrm{integral}: {\int }_{\mathrm{x}=1}^{\infty } \mathrm{f}\left(\mathrm{x}\right) \mathrm{dx} = {\int }_{\mathrm{x}=1}^{\infty }\frac{\mathrm{x}}{{\left(1+{\mathrm{x}}^2\right)}^{\mathrm{p}}}\mathrm{dx}. \\ \mathrm{Therefore}, \\ {\int }_{\mathrm{x}=1}^{\infty } \mathrm{f}\left(\mathrm{x}\right) \mathrm{dx} = \underset{\mathrm{k}\to \infty }{\lim }{\int }_{\mathrm{x}=1}^{\mathrm{k}}\frac{\mathrm{x}}{{\left(1+{\mathrm{x}}^2\right)}^{\mathrm{p}}}\mathrm{dx} \\ =\frac{1}{2}\underset{\mathrm{k}\to \infty }{\lim }{\int }_{\mathrm{u}=2}^{{\mathrm{k}}^2+1}\frac{\mathrm{du}}{{\mathrm{u}}^{\mathrm{p}}} \left(\mathrm{Put} 1+{\mathrm{x}}^2= \mathrm{u}, \Rightarrow 2\mathrm{xdx} = \mathrm{du}\right) \\ = \frac{1}{2}\underset{\mathrm{k}\to \infty }{\lim }{\left[\frac{{\mathrm{u}}^{-\mathrm{p}+1}}{-\mathrm{p}+1}\right]}_2^{{\mathrm{k}}^2+1} \left(\mathrm{Integrating}\right) \\ = \frac{1}{2(-\mathrm{p}+1)}\underset{\mathrm{k}\to \infty }{\lim }{\left[\frac{1}{{\mathrm{u}}^{\mathrm{p}-1}}\right]}_2^{{\mathrm{k}}^2+1} \left(\mathrm{Substitution}\right) \end{gathered}$$

Step 4. Result:

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