Q 48. - Calculus

Question

Find the interval of convergence for power series:  $\sum _{k=0}^{\infty }\frac{k^3}{1.3.5....\left(2k+1\right)}{\left(x+1\right)}^k$

Step-by-Step Solution

Answer

The interval of convergence for power series is $R$.

Step 1. Given information.

The given power series is $\sum _{k=0}^{\infty }\frac{k^3}{1.3.5....\left(2k+1\right)}{\left(x+1\right)}^k$.

Step 2. Find the interval of convergence.

Let us assume $b_k=\frac{k^3}{1.3.5....\left(2k+1\right)}{\left(x+1\right)}^k$ therefore $b_{k+1}=\frac{{\left(k+1\right)}^3}{1.3.5.....\left(2\left(k+1\right)+1\right)}{\left(x+1\right)}^{k+1}$

Ratio for the absolute convergence is 

$$\begin{gathered} \underset{k\to \infty }{\lim }\left|\frac{b_{k+1}}{b_k}\right|=\underset{k\to \infty }{\lim }\left|\frac{{\displaystyle \frac{{\left(k+1\right)}^3}{1.3.5....\left(2\left(k+1\right)+1\right)}}{\left(x+1\right)}^{k+1}}{{\displaystyle \frac{k^3}{1.3.5.....\left(2k+1\right)}}{\left(x+1\right)}^k}\right| \\ =\underset{k\to \infty }{\lim }\left|\frac{1}{2k+3}{\left(\frac{k}{k+3}\right)}^3\left(x+1\right)\right| \end{gathered}$$

So, by ratio test of absolute convergence the series will converge when $\left|x+1\right|<1$.

This implies that

 $-1<x+1<1$

So, $-1<x+1$ and $x+1<1$

$x>-2$ and $x<0$

Step 2. Find the interval of convergence.

Evaluate the series at $x=-2$

$$\begin{gathered} \sum _{k=0}^{\infty }\frac{k^3}{1.3.5....\left(2k+1\right)}{\left(x+1\right)}^k=\sum _{k=0}^{\infty }\frac{k^3}{1.3.5....\left(2k+1\right)}{\left(-2+1\right)}^k \\ \underset{k=0}{\overset{\infty }{=\sum }}\frac{k^3}{1.3.5....\left(2k+1\right)}{\left(-1\right)}^k \end{gathered}$$

Thus, the series will diverge.

Evaluate the series when $x=0$.

$$\begin{gathered} \sum _{k=0}^{\infty }\frac{k^3}{1.3.5....\left(2k+1\right)}{\left(x+1\right)}^k=\sum _{k=0}^{\infty }\frac{k^3}{1.3.5....\left(2k+1\right)}{\left(0+1\right)}^k \\ =\sum _{k=0}^{\infty }\frac{k^3}{1.3.5....\left(2k+1\right)}{\left(1\right)}^k \end{gathered}$$

Thus, the series will diverge.

Therefore the value of for which the series converges is $\left(-2,0\right)$.