Chapter 8: Q 43. (page 670)

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

Short Answer

Expert verified

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

Step by step solution

Step 1. Given information.

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

Step 2. Find the interval of convergence.

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

Ratio for the absolute convergence is

$\lim_{k \to \infty} |\frac{b_{k + 1}}{b_{k}}| = \lim_{k \to \infty} |\frac{\frac{\sqrt{k + 1}}{(k + 1)!} {(4 x + 7)}^{k + 1}}{\frac{\sqrt{k}}{k!} {(4 x + 7)}^{k}}| = \lim_{k \to \infty} |4 x + 7| \sqrt{\frac{k + 1}{k} \frac{1}{k + 1}} = \lim_{k \to \infty} |4 x + 7| \sqrt{\frac{1}{k (k + 1)}}$

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

So, $\lim_{k \to \infty} |4 x + 7| \sqrt{\frac{1}{k (k + 1)}} = 0$ , 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 $R$ .

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

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Find the interval of convergence.

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Find the interval of convergence for power series:∑k=0∞kk!4x+7k

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Find the interval of convergence.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

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