Q 43. - Calculus

Question

Find the interval of convergence for power series: $\sum _{k=0}^{\infty }\frac{\sqrt{k}}{k!}{\left(4x+7\right)}^k$

Step-by-Step Solution

Answer

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

Step 1. Given information.

The given power series is $\sum _{k=0}^{\infty }\frac{\sqrt{k}}{k!}{\left(4x+7\right)}^k$.

Step 2. Find the interval of convergence.

Let us assume $b_k=\frac{\sqrt{k}}{k!}{\left(4x+7\right)}^k$ and $b_{k+1}=\frac{\sqrt{k+1}}{\left(k+1\right)!}{\left(4x+7\right)}^{k+1}$

Ratio for the absolute convergence is 

$$\begin{gathered} \underset{k\to \infty }{\lim }\left|\frac{b_{k+1}}{b_k}\right|=\underset{k\to \infty }{\lim }\left|\frac{{\displaystyle \frac{\sqrt{k+1}}{\left(k+1\right)!}}{\left(4x+7\right)}^{k+1}}{{\displaystyle \frac{\sqrt{k}}{k!}}{\left(4x+7\right)}^k}\right| \\ =\underset{k\to \infty }{\lim }\left|4x+7\right|\sqrt{\frac{k+1}{k}\frac{1}{k+1}} \\ =\underset{k\to \infty }{\lim }\left|4x+7\right|\sqrt{\frac{1}{k(k+1)}} \end{gathered}$$

Now, we evaluate the limit at $k\to \infty$.

So, $\underset{k\to \infty }{\lim }\left|4x+7\right|\sqrt{\frac{1}{k(k+1)}}=0$, that is the value of limit will be zero no matter what value the variable $x$ takes.

Hence by the ratio test the series converges absolutely for every value of $x$.

Therefore, the interval of convergence of the power series is $R$.