Math  /  Calculus

Question4. (15 points) Determine if the series k=111+k2\sum_{k=1}^{\infty} \frac{1}{1+k^{2}} diverge, converges absolutely or conditionally

Studdy Solution

STEP 1

1. We are dealing with the series k=111+k2\sum_{k=1}^{\infty} \frac{1}{1+k^{2}}.
2. We need to determine if the series converges absolutely, converges conditionally, or diverges.
3. We will use standard tests for convergence of series, such as the comparison test and the integral test.

STEP 2

1. Analyze the terms of the series.
2. Apply the comparison test.
3. Determine absolute convergence.
4. Conclude about the convergence of the series.

STEP 3

Analyze the terms of the series:
The terms of the series are 11+k2\frac{1}{1+k^{2}}. As kk increases, k2k^2 becomes much larger than 1, so the terms 11+k2\frac{1}{1+k^{2}} behave similarly to 1k2\frac{1}{k^{2}}.

STEP 4

Apply the comparison test:
We compare 11+k2\frac{1}{1+k^{2}} with 1k2\frac{1}{k^{2}}, a known convergent pp-series with p=2p = 2.
Since 0<11+k2<1k20 < \frac{1}{1+k^{2}} < \frac{1}{k^{2}} for all k1k \geq 1, and k=11k2\sum_{k=1}^{\infty} \frac{1}{k^{2}} converges, by the comparison test, k=111+k2\sum_{k=1}^{\infty} \frac{1}{1+k^{2}} also converges.

STEP 5

Determine absolute convergence:
Since all terms 11+k2\frac{1}{1+k^{2}} are positive, absolute convergence is equivalent to convergence in this case. The series converges absolutely.

STEP 6

Conclude about the convergence of the series:
The series k=111+k2\sum_{k=1}^{\infty} \frac{1}{1+k^{2}} converges absolutely.
The series converges absolutely.

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