Chapter 8: Q. 3 (page 679)

Show that the series: $\ln (2) + \sum_{k = 1}^{\infty} \frac{{(- 1)}^{k - 1}}{k 2^{k}} {(x - 2)}^{k}$
from Example 3 diverges when x = 0 and converges conditionally when x = 4.

Short Answer

Expert verified

$Series diverges if x =0 since the harmonic series diverges . Series converges conditionally if x =4.$

Step by step solution

Step 1. Given information is:

$\ln (2) + \sum_{k = 1}^{\infty} \frac{{(- 1)}^{k - 1}}{k 2^{k}} {(x - 2)}^{k}$

Step 2. Check Divergence

$Assume that x = 0; f (x = 0) = \ln (2) + \sum_{k = 1}^{\infty} \frac{{(- 1)}^{k - 1}}{k 2^{k}} {(0 - 2)}^{k} f (x = 0) = \ln (2) - \sum_{k = 1}^{\infty} \frac{1}{k} Therefore, series diverges if x = 0 since the harmonic series diverges .$

Step 3. Check Convergence

$Assume that x = 4; f (x = 4) = \ln (2) + \sum_{k = 1}^{\infty} \frac{{(- 1)}^{k - 1}}{k 2^{k}} {(4 - 2)}^{k} f (x = 4) = \ln (2) + \sum_{k = 1}^{\infty} \frac{{(- 1)}^{k - 1}}{k 2^{k}} {(2)}^{k} Therefore, series converges conditionally if x = 4 .$

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Show that the series: $\ln (2) + \sum_{k = 1}^{\infty} \frac{{(- 1)}^{k - 1}}{k 2^{k}} {(x - 2)}^{k}$
from Example 3 diverges when x = 0 and converges conditionally when x = 4.

Short Answer

Step by step solution

Step 1. Given information is:

Step 2. Check Divergence

Step 3. Check Convergence

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Show that the series:ln2+∑k=1∞-1k-1k2kx-2kfrom Example 3 diverges when x = 0 and converges conditionally when x = 4.

Short Answer

Step by step solution

Step 1. Given information is:

Step 2. Check Divergence

Step 3. Check Convergence

One App. One Place for Learning.

Most popular questions from this chapter

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Show that the series: $\ln (2) + \sum_{k = 1}^{\infty} \frac{{(- 1)}^{k - 1}}{k 2^{k}} {(x - 2)}^{k}$
from Example 3 diverges when x = 0 and converges conditionally when x = 4.