91Ó°ÊÓ

Givenfunction:$f\left(x\right)=\underset{i=1}{\overset{∞}{∑}}\frac{(-1{)}^d}{k}(x+5{)}^k$

Let us first assume

$$\begin{gathered} b_k=\frac{(-1{)}^k}{k}(x+5{)}^k \\ ⇒b_{k+1}=\frac{(-1{)}^{k+1}}{k+1}(x+5{)}^{k+1} \end{gathered}$$

Therefore,

$$\begin{gathered} \underset{k→∞}{\lim }\left|\frac{b_{k+1}}{b_k}\right|=\underset{k→∞}{\lim }âˆ\pounds \frac{\frac{(-1{)}^{k+1}}{k+1}(x+5{)}^{k+1}}{(-1{)}^k}(x+5{)}^k \\ =\underset{k→∞}{\lim }\frac{-k}{(k+1)}|x+5| \end{gathered}$$

Now, by the ratio test for absolute convergence, the series will converge only when $|x+5|<1$

Therefore,$x∈(-6,-4)$

Now, we check the series at the end points

So, when$x=-6$

$$\begin{gathered} \underset{k=1}{\overset{∞}{∑}}\frac{(-1{)}^k}{k}(-6+5{)}^k=\underset{k=1}{\overset{∞}{∑}}\frac{(-1{)}^{2k}}{k} \\ =\underset{k=1}{\overset{∞}{∑}}\frac{1}{k} \end{gathered}$$

This series will diverge.

When$x=4$

$\underset{k=1}{\overset{∞}{∑}}\frac{(-1{)}^k}{k}(4+5{)}^k=\underset{k=1}{\overset{∞}{∑}}\frac{(-1{)}^k{9}^k}{k}$

This series will converge.

Therefore, the interval of convergence of the power series is$\mathbf{(}\mathbf{-}\mathbf{6}\mathbf{,}\mathbf{-}\mathbf{4}\mathbf{]}$

Givenfunction:$f\left(x\right)=\underset{i=1}{\overset{∞}{∑}}\frac{(-1{)}^d}{k}(x+5{)}^k$

Derivative of the function$f\left(x\right)$

Therefore,

$$\begin{gathered} f^{\text{'}}\left(x\right)=\frac{d}{dx}\left[\underset{i=1}{\overset{∞}{∑}}\frac{(-1{)}^k}{k}(x+5{)}^k\right] \\ =\underset{i=1}{\overset{∞}{∑}}\frac{(-1{)}^d}{k}\frac{d}{dx}(x+5{)}^k \\ =\underset{i=1}{\overset{∞}{∑}}\frac{(-1{)}^k}{k}k(x+5{)}^{k-1} \\ =\underset{i=1}{\overset{∞}{∑}}(-1{)}^k(x+5{)}^{k-1} \end{gathered}$$

Now, we change the index in the final step. So, the power series in $x-x_0$for$f\text{'}\left(x\right)$ is

$f^{\text{'}}\left(x\right)=\underset{\mathbf{t}\mathbf{=}\mathbf{0}}{\overset{\mathbf{∞}}{\mathbf{∑}}}\mathbf{(}\mathbf{-}\mathbf{1}{\mathbf{)}}^{\mathbf{k}\mathbf{+}\mathbf{1}}\mathbf{(}\mathit{x}\mathbf{+}\mathbf{5}{\mathbf{)}}^{\mathbf{k}}$

Givenfunction:$f\left(x\right)=\underset{i=1}{\overset{∞}{∑}}\frac{(-1{)}^d}{k}(x+5{)}^k$

Also, to find the power series in $x-x_0$for $F$, let us integrate the function$f\left(x\right)$from $x_0$to$x$

Therefore,

$$\begin{gathered} F\left(x\right)={∫}_k^x\left[\underset{i=1}{\overset{k}{∑}}\frac{(-1{)}^k}{k}(t+5{)}^k\right]dt \\ =\underset{k=1}{\overset{5}{∑}}\frac{(-1{)}^k}{k}{∫}_{5_0}^k\left[(t+5{)}^k\right]dt \\ =\underset{k=1}{\overset{k}{∑}}\frac{(-1{)}^k}{k}{\left[\frac{(t+5{)}^{k+1}}{k+1}\right]}_k^k \end{gathered}$$

Thus,

$F\left(x\right)=\underset{k=1}{\overset{∞}{∑}}\frac{(-1{)}^2}{k(k+1)}(x+5{)}^{k+1}$

Now, we change the index in the final step

So, the power series in $x-x_0$for $F\left(x\right)={∫}_{x_0}^{x}f\left(t\right)dt$is

$F\left(x\right)=\underset{\mathbf{k}\mathbf{=}\mathbf{2}}{\overset{\mathbf{∞}}{\mathbf{∑}}}\frac{\mathbf{(}\mathbf{-}\mathbf{1}{\mathbf{)}}^{\mathbf{k}\mathbf{-}\mathbf{1}}}{\mathbf{k}\mathbf{(}\mathbf{k}\mathbf{-}\mathbf{1}\mathbf{)}}\mathbf{(}\mathit{x}\mathbf{+}\mathbf{5}{\mathbf{)}}^{\mathbf{k}}$