Find the interval of convergence for power series:$\sum _{k=1}^{\infty }\frac{{\left(-1\right)}^k}{k+1}{x}^k$

The given power series is$\sum _{k=1}^{\infty }\frac{{\left(-1\right)}^k}{k+1}{x}^k$.

Let us assume $b_k=\frac{{\left(-1\right)}^k}{k+1}{x}^k$and $b_{k+1}=\frac{{\left(-1\right)}^k}{k+1+1}{x}^{k+1}$

Ratio for the absolute convergence is

$$\begin{gathered} \underset{k\to \infty }{\lim }\left|\frac{b_{k+1}}{b_k}\right|=\underset{k\to \infty }{\lim }\left|\frac{{\displaystyle \frac{{\left(-1\right)}^{k+1}}{k+2}}{x}^{k+1}}{{\displaystyle \frac{{\left(-1\right)}^k}{k+1}}{x}^k}\right| \\ =\underset{k\to \infty }{\lim }\left|\frac{-1(k+1)}{k+2}x\right| \\ =\underset{k\to \infty }{\lim }\left|x\right|\frac{(k+1)}{k+2} \end{gathered}$$

Here the limit is $\left|x\right|$.

So, by the ratio test of absolute convergence, we know that the series will converge absolutely, when$\left|x\right|<1$that is$-1<x<1$.

Now, since the intervals are finite so we analyze the behavior of the series at the endpoints. So, when$x=-1$

$$\begin{gathered} \sum _{k=1}^{\infty }\frac{{\left(-1\right)}^k}{k+1}{x}^k=\sum _{k=1}^{\infty }\frac{{\left(-1\right)}^k{\left(-1\right)}^k}{k+1} \\ =\sum _{k=1}^{\infty }\frac{{\left(-1\right)}^{2k}}{k+1} \end{gathered}$$

Implies that $\sum _{k=1}^{\infty }\frac{{\left(-1\right)}^k}{k+1}{x}^k=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....$

The result is just a constant multiple of the harmonic series, which diverges.

So, when$x=1$

$$\begin{gathered} \sum _{k=1}^{\infty }\frac{{\left(-1\right)}^k}{k+1}{x}^k=\sum _{k=1}^{\infty }\frac{{\left(-1\right)}^k{\left(1\right)}^k}{k+1} \\ =\sum _{k=1}^{\infty }\frac{{\left(-1\right)}^k}{k+1} \end{gathered}$$

The result is the alternating multiple of the harmonic series, which converges conditionally.

Therefore, the interval of convergence of the power series$\sum _{k=1}^{\infty }\frac{{\left(-1\right)}^k}{k+1}{x}^k$is$(-1,1]$.