Q 30. - Calculus

Question

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

Step-by-Step Solution

Answer

The interval of convergence for power series is $(0,2]$.

Step 1. Given information.

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

Step 2. Find the interval of convergence.

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

Ratio for the absolute convergence is 

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

Here the limit is $\left|x-1\right|$So, by the ratio test of absolute convergence, we know that series will converge absolutely when $\left|x-1\right|<1$ that is $0<x<2$.

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 $x=0$

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

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

So, when $x=2$

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

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

Therefore, the interval of convergence of the power series is $(0,2]$.