Question 8.5.4: Using the Ratio Test ∑^∞k=1 (−1)^k k/2k for convergence.

Calculus: Early Transcendental Functions

Using the Ratio Test

Test $\sum_{k=1}^{\infty} \frac{(-1)^k k}{2^k}$ for convergence.

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The graph of the first 20 partial sums of the series of absolute values, $\sum_{k=1}^{\infty} \frac{k}{2^k}$ , seen in Figure 8.36, suggests that the series of absolute values converges to about 2. From
the Ratio Test, we have

\lim _{k \rightarrow \infty}\left|\frac{a_{k+1}}{a_k}\right|=\lim _{k \rightarrow \infty} \frac{\frac{k+1}{2^{k+1}}}{\frac{k}{2^k}}=\lim _{k \rightarrow \infty} \frac{k+1}{2^{k+1}} \frac{2^k}{k}=\frac{1}{2} \lim _{k \rightarrow \infty} \frac{k+1}{k}=\frac{1}{2}<1 \quad\stackrel{\text { Since }}{2^{k+1}}=2^k \cdot 2^1 .

and so, the series converges absolutely, as expected from Figure 8.36.

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