Q 27. - Calculus

Question

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

Step-by-Step Solution

Answer

The interval of convergence for power series is $[-3,-1)$.

Step 1. Given information.

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

Step 2. Find the interval of convergence.

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

Ratio for the absolute convergence is 

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

Here, the limit is $\left|x+2\right|$. So, by the ratio test of absolute convergence. We know that the series will converge absolutely. When $\left|x+2\right|<1$that is $-3<x<-1$

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 $x=-3$

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

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

So, when $x=-1$

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

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

Therefore, the interval of convergence of the power series is $[-3,-1)$.