Q 53. - Calculus

Question

Explain why the series is not a power series in $x-x_0$. Then use the ratio test for absolute convergence to find the values of $x$ for which the given series converge.

$\sum _{k=0}^{\infty }{\left(\frac{1}{x+1}\right)}^k$

Step-by-Step Solution

Answer

The value of $x$ for which the series$\sum _{k=0}^{\infty }{\left(\frac{1}{x+1}\right)}^k$ converges when $x>-2$ and $x>0$.

Step 1. Given information.

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

Step 2. Find the values of x for which the given series converge.

This series can be written as $\sum _{k=0}^{\infty }{\left(\frac{1}{x+1}\right)}^k=\sum _{k=0}^{\infty }{\left(x+1\right)}^{-k}$

or can be written as $\sum _{k=0}^{\infty }{\left(\frac{1}{x+1}\right)}^k=\sum _{k=0}^{\infty }{\left(x-\left(-1\right)\right)}^k$

Since the power of $x+1$ are negative so the series is not power series.

Now, $b_k={\left(x+1\right)}^{-k}$ and $b_{k+1}={\left(x+1\right)}^{-\left(k+1\right)}$

Ratio for the absolute convergence is 

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

So, by the ratio test of absolute convergence, the series will converge when 

$\left|\frac{1}{\left(x+1\right)}\right|<1$

So, $\left|\left(x+1\right)\right|>1$

This implies that

$-\left(x+1\right)<1$and $1<x+1$

$x>-2$ and $x>0$

Therefore, the value of  $x$ for which the series $\sum _{k=0}^{\infty }{\left(\frac{1}{x+1}\right)}^k$  converges when $x>-2$ and $x>0$