Question

Find the interval of convergence of each of the following power series; be sure to investigate the endpoints of the interval in each case ${\mathbf{\sum }}_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}\frac{\mathbf{(}\mathbf{-}\mathbf{1}{\mathbf{)}}^{\mathbf{n}}{\mathbf{x}}^{\mathbf{n}}}{\mathbf{n}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}}$.

Short answer

The interval of convergence of the power series is $-1\le \mathrm{x}\le 1$.

Step-by-step solution

Determine the method to get the interval

Let us consider the power series

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

Determine the interval of convergence of the given power series.

Start using the ratio test; follow the steps as mentioned below:

First,find.${\mathbf{\rho }}_{\mathbf{n}}\mathbf{=}\left|\frac{a_{n+1}}{a_n}\right|\mathbf{.}$

Second,$\mathbf{\rho }\mathbf{=}\underset{\mathbf{n}\mathbf{\to }\mathbf{\infty }}{\mathbf{lim}\mathbf{}{\mathbf{\rho }}_{\mathbf{n}}}$.

Find pn and p

If$\rho <1$,the series converges, if$\rho >1$the series diverges.

So,we divide term$(n+1)$by term n and take the absolute value of this

Ratio to get the value of${\rho }_n$ .

Further, take the limit of ${\mathrm{\rho }}_{\mathrm{n}}\mathrm{as}\mathrm{n}\to \infty \mathrm{to}\mathrm{get}\mathrm{\rho }$.

$$\begin{gathered} {\mathrm{p}}_{\mathrm{n}}=\left|\frac{{\mathrm{a}}_{\mathrm{n}+1}}{{\mathrm{a}}_{\mathrm{n}}}\right| \\ =\left|\frac{{\displaystyle \frac{{\mathrm{x}}^{\mathrm{n}+1}}{\left(\mathrm{n}+2\right)\left(\mathrm{n}+1\right)}}}{{\displaystyle \frac{{\mathrm{x}}^{\mathrm{n}}}{\mathrm{n}\left(\mathrm{n}+1\right)}}}\right| \\ =\left|\frac{{\mathrm{x}}^{\mathrm{n}+1}\mathrm{n}\left(\mathrm{n}+1\right)}{\left(\mathrm{n}+2\right)\left(\mathrm{n}+1\right){\mathrm{x}}^{\mathrm{n}}}\right| \\ =\frac{\mathrm{nx}}{\mathrm{n}+2} \\ \mathrm{p}=\underset{\mathrm{n}\to \infty }{\lim }\left|\frac{\mathrm{nx}}{\mathrm{n}+2}\right| \\ =\left|\mathrm{x}\right| \end{gathered}$$

Check the conditions for convergence and divergence

Now, the series converges for$\rho <1$which is possible when$\left|x\right|<1$ .

The series will diverge for $\left|x\right|>1$.

Hence, we can say that the series converges when$-1<\mathrm{x}<1$.

Next, to get the endpoints of the convergence interval, when$x=1$ , the series is :

$\sum _{\mathrm{n}=1}^{\infty }\frac{(-1{)}^{\mathrm{n}}}{\mathrm{n}(\mathrm{n}+1)}$this is a convergent alternating series.

Also, for$x=-1$ , the series is :

role="math" localid="1658815517961" $\sum _{\mathrm{n}=1}^{\infty }\frac{(1{)}^{\mathrm{n}}}{\mathrm{n}(\mathrm{n}+1)}$this is also a convergent series.

It converges by comparing it with the convergent series$\sum _{\mathrm{n}=1}^{\infty }\frac{1}{{\mathrm{n}}^2}$.

Finally, we conclude that the interval of convergence of the power series is$-1\le x\le 1$