Math  /  Calculus

Question8) Which of the following series is conditionally convergent? A) k=1(1)k1ln(k+1)\sum_{k=1}^{\infty}(-1)^{k} \frac{1}{\ln (k+1)} B) k=1(1)k1k2+1\sum_{k=1}^{\infty}(-1)^{k} \frac{1}{k^{2}+1} C) k=1(1)kk2k+1\sum_{k=1}^{\infty}(-1)^{k} \frac{k}{2 k+1} D) k=1(1)k(2k+1k+1)k\sum_{k=1}^{\infty}(-1)^{k}\left(\frac{2 k+1}{k+1}\right)^{k}

Studdy Solution

STEP 1

1. A series k=1ak\sum_{k=1}^{\infty} a_k is conditionally convergent if it converges, but the series of absolute values k=1ak\sum_{k=1}^{\infty} |a_k| diverges.
2. We will use the Alternating Series Test (Leibniz Test) to check if the series converges.
3. We will check the absolute convergence by applying relevant convergence tests to the absolute series k=1ak\sum_{k=1}^{\infty} |a_k|.

STEP 2

1. Verify the alternating series condition for each option.
2. Apply the Alternating Series Test for convergence.
3. Check for absolute convergence using appropriate tests.
4. Determine if the series is conditionally convergent.

STEP 3

Examine Option A: k=1(1)k1ln(k+1)\sum_{k=1}^{\infty} (-1)^{k} \frac{1}{\ln(k+1)}
Check if 1ln(k+1)\frac{1}{\ln(k+1)} decreases to 0 as kk \to \infty.
limk1ln(k+1)=0 \lim_{k \to \infty} \frac{1}{\ln(k+1)} = 0
Since ln(k+1)\ln(k+1) increases without bound, 1ln(k+1)\frac{1}{\ln(k+1)} decreases to 0.

STEP 4

Apply the Alternating Series Test for Option A.
Check if 1ln(k+1)\frac{1}{\ln(k+1)} is monotonically decreasing.
Since ln(k+1)\ln(k+1) is increasing, 1ln(k+1)\frac{1}{\ln(k+1)} is decreasing. By the Alternating Series Test, k=1(1)k1ln(k+1)\sum_{k=1}^{\infty} (-1)^{k} \frac{1}{\ln(k+1)} converges.

STEP 5

Check absolute convergence for Option A by examining k=11ln(k+1)=k=11ln(k+1)\sum_{k=1}^{\infty} \left| \frac{1}{\ln(k+1)} \right| = \sum_{k=1}^{\infty} \frac{1}{\ln(k+1)}.
Use the Integral Test to check convergence.
Consider the integral:
11ln(x+1)dx \int_{1}^{\infty} \frac{1}{\ln(x+1)} \, dx
This integral diverges because the integrand is a slowly decreasing function.

STEP 6

Since k=1(1)k1ln(k+1)\sum_{k=1}^{\infty} (-1)^{k} \frac{1}{\ln(k+1)} converges but k=11ln(k+1)\sum_{k=1}^{\infty} \frac{1}{\ln(k+1)} diverges, Option A is conditionally convergent.

STEP 7

Examine Option B: k=1(1)k1k2+1\sum_{k=1}^{\infty} (-1)^{k} \frac{1}{k^{2}+1}
Check if 1k2+1\frac{1}{k^{2}+1} decreases to 0 as kk \to \infty.
limk1k2+1=0 \lim_{k \to \infty} \frac{1}{k^{2}+1} = 0
Since k2+1k^2+1 increases without bound, 1k2+1\frac{1}{k^{2}+1} decreases to 0.

STEP 8

Apply the Alternating Series Test for Option B.
Check if 1k2+1\frac{1}{k^{2}+1} is monotonically decreasing.
Since k2+1k^2+1 is increasing, 1k2+1\frac{1}{k^{2}+1} is decreasing. By the Alternating Series Test, k=1(1)k1k2+1\sum_{k=1}^{\infty} (-1)^{k} \frac{1}{k^{2}+1} converges.

STEP 9

Check absolute convergence for Option B by examining k=11k2+1=k=11k2+1\sum_{k=1}^{\infty} \left| \frac{1}{k^{2}+1} \right| = \sum_{k=1}^{\infty} \frac{1}{k^{2}+1}.
Compare with k=11k2\sum_{k=1}^{\infty} \frac{1}{k^2}.
Since 1k2+11k2\frac{1}{k^{2}+1} \leq \frac{1}{k^2} and k=11k2\sum_{k=1}^{\infty} \frac{1}{k^2} converges, k=11k2+1\sum_{k=1}^{\infty} \frac{1}{k^{2}+1} also converges by the Comparison Test.

STEP 10

Since both k=1(1)k1k2+1\sum_{k=1}^{\infty} (-1)^{k} \frac{1}{k^{2}+1} and k=11k2+1\sum_{k=1}^{\infty} \frac{1}{k^{2}+1} converge, Option B is absolutely convergent and not conditionally convergent.

STEP 11

Examine Option C: k=1(1)kk2k+1\sum_{k=1}^{\infty} (-1)^{k} \frac{k}{2k+1}
Check if k2k+1\frac{k}{2k+1} decreases to 0 as kk \to \infty.
limkk2k+1=120 \lim_{k \to \infty} \frac{k}{2k+1} = \frac{1}{2} \neq 0
Since k2k+1\frac{k}{2k+1} does not decrease to 0, the series does not satisfy the condition for the Alternating Series Test and does not converge.

STEP 12

Examine Option D: k=1(1)k(2k+1k+1)k\sum_{k=1}^{\infty} (-1)^{k} \left(\frac{2k+1}{k+1}\right)^{k}
Check if (2k+1k+1)k\left(\frac{2k+1}{k+1}\right)^{k} decreases to 0 as kk \to \infty.
Consider the limit:
limk(2k+1k+1)k=limk(2+1k+1)k= \lim_{k \to \infty} \left(\frac{2k+1}{k+1}\right)^{k} = \lim_{k \to \infty} \left(2 + \frac{1}{k+1}\right)^{k} = \infty
Since (2k+1k+1)k\left(\frac{2k+1}{k+1}\right)^{k} does not decrease to 0, the series does not satisfy the condition for the Alternating Series Test and does not converge.

STEP 13

After evaluating all options, the only series that is conditionally convergent is Option A.
Thus, the correct answer is:
Option A: k=1(1)k1ln(k+1) \text{Option A: } \sum_{k=1}^{\infty}(-1)^{k} \frac{1}{\ln (k+1)}

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