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Chapter 9: Problem 21

Use the Root Test to determine whether the following series converge. $$\sum_{k=1}^{\infty} \frac{k^{2}}{2^{k}}$$

Short Answer

Expert verified
Based on the Root Test, the series \(\sum_{k=1}^{\infty} \frac{k^{2}}{2^{k}}\) converges, since the limit of the n-th root of the absolute value of the general term is \(\frac{1}{2}\), which is less than 1.

Step by step solution

01

Write the general term and find its absolute value

The given series is \(\sum_{k=1}^{\infty} \frac{k^{2}}{2^k}\). The general term of the series is: $$a_k = \frac{k^{2}}{2^k}$$ Since the term is already positive, the absolute value remains the same: $$|a_k| = \frac{k^{2}}{2^k}$$
02

Apply the Root Test

Now, we need to find the limit of the \(n\)-th root of the absolute value of \(a_k\) as \(k\) goes to infinity. So, we are looking for the following limit: $$\lim_{k \to \infty} \sqrt[k]{|a_k|} = \lim_{k \to \infty} \sqrt[k]{\frac{k^{2}}{2^{k}}}$$
03

Simplify the expression and find the limit

Using the properties of exponents, we can rewrite the expression within the limit as: $$\lim_{k \to \infty} \sqrt[k]{\frac{k^{2}}{2^{k}}} = \lim_{k \to \infty} \frac{\sqrt[k]{k^2}}{\sqrt[k]{2^{k}}}$$ Now, rewrite the expression as: $$\lim_{k \to \infty} \frac{k^{\frac{2}{k}}}{2}$$ Since the numerator converges to 1 as \(k\) goes to infinity, we have: $$\lim_{k \to \infty} \frac{k^{\frac{2}{k}}}{2} = \frac{1}{2}$$
04

Determine the convergence of the series

As the limit in the Root Test is less than 1 (in our case, it is \(\frac{1}{2}<1\)), we can conclude that the given series converges: $$\sum_{k=1}^{\infty} \frac{k^{2}}{2^{k}} \text{ converges}.$$

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