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Question
Math
Posted 9 months ago
n=1(1)n+1nn2+1
\sum_{n=1}^{\infty}(-1)^{n+1} \frac{n}{n^{2}+1}


Does the series converge conditionally, converge absolutely, or diverge?
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Answer from Sia
Posted 9 months ago
Solution by Steps
step 1
To determine the convergence of the series, we first consider the alternating series test
step 2
The alternating series test states that if the absolute value of the terms an=nn2+1 a_n = \frac{n}{n^2+1} is decreasing and limnan=0 \lim_{n \to \infty} a_n = 0 , then the series n=1(1)n+1an \sum_{n=1}^{\infty} (-1)^{n+1} a_n converges
step 3
We show that an a_n is decreasing by showing that an+1<an a_{n+1} < a_n for all n n
step 4
We calculate the limit limnan=limnnn2+1 \lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{n}{n^2+1} and find that it equals 0
step 5
Since both conditions of the alternating series test are met, the series converges
step 6
To determine if the series converges absolutely, we consider the series n=1(1)n+1an=n=1nn2+1 \sum_{n=1}^{\infty} |(-1)^{n+1} a_n| = \sum_{n=1}^{\infty} \frac{n}{n^2+1}
step 7
We use the comparison test, comparing nn2+1 \frac{n}{n^2+1} with 1n \frac{1}{n} , and note that nn2+1 \frac{n}{n^2+1} is smaller for all n1 n \geq 1
step 8
Since n=11n \sum_{n=1}^{\infty} \frac{1}{n} diverges (harmonic series), and nn2+1 \frac{n}{n^2+1} is smaller, we cannot conclude that n=1nn2+1 \sum_{n=1}^{\infty} \frac{n}{n^2+1} converges
step 9
Therefore, we cannot say that the series converges absolutely
step 10
The series n=1(1)n+1nn2+1 \sum_{n=1}^{\infty}(-1)^{n+1} \frac{n}{n^{2}+1} converges conditionally based on the alternating series test
Answer
The series converges conditionally.
Key Concept
Alternating Series Test and Absolute Convergence
Explanation
The series passes the alternating series test and thus converges, but it does not converge absolutely because the absolute series does not pass the comparison test with the harmonic series.

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