Q 26. - Calculus

Question

Find the interval of convergence for power series: $\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{2k+1}{x}^{2k+1 }$

Step-by-Step Solution

Answer

The interval of convergence for power series is $[-1,1]$.

Step 1. Given information.

The given power series is  $\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{2k+1}{x}^{2k+1 }$

Step 2. Find the interval of convergence.

Let us assume $b_k=\frac{{\left(-1\right)}^k}{2k+1}{x}^{2k+1}$and

 $$\begin{gathered} b_{k+1}=\frac{{\left(-1\right)}^{k+1}}{2\left(k+1\right)+1}{x}^{2\left(k+1\right)+1} \\ {b}_{k+1}=\frac{{\left(-1\right)}^{k+1}}{2k+3}{x}^{2k+3} \end{gathered}$$

Ratio for the absolute convergence is 

$$\begin{gathered} \underset{k\to \infty }{\lim }\left|\frac{b_{k+1}}{b_k}\right|=\underset{k\to \infty }{\lim }\left|\frac{{\displaystyle \frac{{\left(-1\right)}^{k+1}}{2k+3}}{x}^{2k+3}}{{\displaystyle \frac{{\left(-1\right)}^k}{2k+1}}{x}^{2k+1}}\right| \\ =\underset{k\to \infty }{\lim }\left|\frac{-1\left(2k+1\right)}{2k+3}{x}^2\right| \\ =\underset{k\to \infty }{\lim }\left|x^2\right|\frac{2k+1}{2k+3} \end{gathered}$$

Here the limit is $\left|x^2\right|$. So, by the ratio test of absolute convergence, we know that series will converge absolutely when $\left|x^2\right|<1$ that is $-1<x<1$. 

Step 3. Find the interval of convergence.

Now, since the intervals are finite so we analyse the behavior of the series at the endpoints.

So, when $x=-1$

$$\begin{gathered} \sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{2k+1}{x}^{2k+1}=\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k{\left(-1\right)}^{2k+1}}{2k+1} \\ =\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^{3k+1}}{2k+1} \end{gathered}$$

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

So, when $x=1$

 $$\begin{gathered} \sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{2k+1}{x}^{2k+1}=\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k{\left(1\right)}^{2k+1}}{2k+1} \\ =\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{2k+1} \end{gathered}$$

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

Therefore, the interval of convergence of the power series is $\left[-1,1\right]$