Chapter 8: Q 22. (page 670)

Find the interval of convergence for power series: $\sum_{k = 0}^{\infty} \frac{{(- 2)}^{k}}{k!} x^{k}$

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 $\sum_{k = 0}^{\infty} \frac{{(- 2)}^{k}}{k!} x^{k}$ .

Step 2. Find the interval of convergence.

Let us assume $b_{k} = \frac{{(- 2)}^{k}}{k!} x^{k}$ and role="math" localid="1649237026776" $b_{k + 1} = \frac{{(- 2)}^{k + 1}}{(k + 1)!} x^{k + 1}$

Ratio for the absolute convergence is

$\lim_{k \to \infty} |\frac{b_{k + 1}}{b_{k}}| = \lim_{k \to \infty} |\frac{\frac{{(- 2)}^{k + 1}}{(k + 1)!} x^{k + 1}}{\frac{{(- 2)}^{k}}{k!} x^{k}}| \lim_{k \to \infty} |x| \frac{- 2}{k + 1}$

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

So, $\lim_{k \to \infty} |x| \frac{1}{k + 1} = 0$ that is the value of limit will be zero no matter what value the variable $x$ takes.

By ratio test, the series converges absolutely for every value of $x$ .

Therefore, the interval of convergence of the power series $\sum_{k = 0}^{\infty} \frac{{(- 2)}^{k}}{k!} x^{k}$ is $R$ .

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Find the interval of convergence for power series: $\sum_{k = 0}^{\infty} \frac{{(- 2)}^{k}}{k!} x^{k}$

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∞-2kk!xk

Short Answer

Step by step solution

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One App. One Place for Learning.

Most popular questions from this chapter

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Find the interval of convergence for power series: $\sum_{k = 0}^{\infty} \frac{{(- 2)}^{k}}{k!} x^{k}$