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Expert-verified Found in: Page 653 ### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861 # Find the values of p that make the series converge absolutely, the values that make the series converge conditionally, and the values that make the series diverge. $\sum_{k=0}^{\infty}(-1)^{k}k^{p}e^{-k^{2}}$

The values of p that the series converges absolutely is p < 2, the series converge conditionally is p < 1 and the series diverges is p>=1

See the step by step solution

## Step 1.Given Information

The given series is $\sum_{k=0}^{\infty}(-1)^{k}k^{p}e^{-k^{2}}$

## Step 2. Calculation

For the series to converge absolutely, the absolute value of each term must approach 0 as k approaches infinity. This means that

$\sum_{k=0}^{\infty}(-1)^{k}k^{p}e^{-k^{2}}$ must approach to 0 as k approaches infinity.

The absolute value of (-1)^k is always 1, so we can ignore it. The absolute value of k^p approaches infinity as k approaches infinity if p is greater than or equal to 0, and approaches 0 if p is less than 0. The absolute value of e^(-k^2) approaches 0 as k approaches infinity, since the exponent is negative.

Therefore, for the series to converge absolutely, we need p < 2.

For the series to converge conditionally, we need the absolute value of each term to decrease monotonically and approach 0, but the signs of the terms can alternate. This means that we need the exponent on the e term to be greater than or equal to 1.

In the general term of the series $\sum_{k=0}^{\infty}(-1)^{k}k^{p}e^{-k^{2}}$

The absolute value of the e term is e^(-k^2), which approaches 0 as k approaches infinity since the exponent is negative. The absolute value of k^p also approaches 0 as k approaches infinity if p is less than 1, since k is an integer and p is a real number.

Therefore, for the series to converge conditionally, we need p < 1, so that the absolute value of each term decreases monotonically and approaches 0 as k approaches infinity.

For the series to diverge, the absolute value of at least one term must not approach 0 as k approaches infinity. This means that $\sum_{k=0}^{\infty}(-1)^{k}k^{p}e^{-k^{2}}$ must not approach 0 as k approaches infinity.

The absolute value of k^p also approaches 0 as k approaches infinity if p is less than 1, since k is an integer and p is a real number.

Therefore, for the series to diverge, we need p >= 1, so that the absolute value of at least one term does not approach 0 as k approaches infinity. ### Want to see more solutions like these? 