Question

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

Short answer

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

Step-by-step solution

Step 1. Given information. 

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

Step 2. Find the interval of convergence.  

Here we use modified root test for the series. This is reasonable choice for the series because the factors of the terms of the series involves $k^{th}$power.

Let us take $b_k=\frac{1}{k^k}{\left(x-3\right)}^k$

Thus, $$\begin{gathered} \underset{k\to \infty }{\lim }\sqrt[k]{\left|b_k\right|}=\underset{k\to \infty }{\lim }\sqrt[k]{\left|\frac{1}{k^k}{\left(x-3\right)}^k\right|} \\ =\underset{k\to \infty }{\lim }\frac{1}{k}\left|x-3\right| \end{gathered}$$

Now, we evaluating the preceding limit as $k\to \infty$.No matter what the variable $x$takes on the limit is zero.

That is $\underset{k\to \infty }{\lim }\frac{1}{k}\left|x-3\right|=0$

Therefore, by the modified root test the series converges absolutely for every value of $x$.

Thus, the interval of the convergence is$R$.