We are asked to determine whether the series $\sum_{n=1}^{\infty} (-1)^{n+1} \frac{n}{n^2+1}$ converges or diverges. Furthermore, if it converges, we want to determine if it converges absolutely or conditionally.

AnalysisSeriesConvergenceDivergenceAlternating Series TestLimit Comparison TestAbsolute ConvergenceConditional Convergence
2025/3/12

1. Problem Description

We are asked to determine whether the series n=1(1)n+1nn2+1\sum_{n=1}^{\infty} (-1)^{n+1} \frac{n}{n^2+1} converges or diverges. Furthermore, if it converges, we want to determine if it converges absolutely or conditionally.

2. Solution Steps

Let an=nn2+1a_n = \frac{n}{n^2+1}.
First, let us examine if the series n=1(1)n+1nn2+1=n=1nn2+1\sum_{n=1}^{\infty} |(-1)^{n+1} \frac{n}{n^2+1}| = \sum_{n=1}^{\infty} \frac{n}{n^2+1} converges absolutely.
We use the Limit Comparison Test with bn=1nb_n = \frac{1}{n}.
limnanbn=limnnn2+11n=limnn2n2+1=limn11+1n2=1\lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{\frac{n}{n^2+1}}{\frac{1}{n}} = \lim_{n \to \infty} \frac{n^2}{n^2+1} = \lim_{n \to \infty} \frac{1}{1+\frac{1}{n^2}} = 1.
Since n=11n\sum_{n=1}^{\infty} \frac{1}{n} is a divergent harmonic series, by the Limit Comparison Test, n=1nn2+1\sum_{n=1}^{\infty} \frac{n}{n^2+1} also diverges.
Therefore, the series n=1(1)n+1nn2+1\sum_{n=1}^{\infty} (-1)^{n+1} \frac{n}{n^2+1} does not converge absolutely.
Now, let us check if the alternating series n=1(1)n+1nn2+1\sum_{n=1}^{\infty} (-1)^{n+1} \frac{n}{n^2+1} converges conditionally using the Alternating Series Test.
For the alternating series test to apply, we must show that an=nn2+1a_n = \frac{n}{n^2+1} is decreasing and limnan=0\lim_{n \to \infty} a_n = 0.
limnnn2+1=limn1n1+1n2=01+0=0\lim_{n \to \infty} \frac{n}{n^2+1} = \lim_{n \to \infty} \frac{\frac{1}{n}}{1+\frac{1}{n^2}} = \frac{0}{1+0} = 0.
To check if ana_n is decreasing, we can consider the derivative of the continuous function f(x)=xx2+1f(x) = \frac{x}{x^2+1}.
f(x)=(x2+1)(1)x(2x)(x2+1)2=x2+12x2(x2+1)2=1x2(x2+1)2f'(x) = \frac{(x^2+1)(1) - x(2x)}{(x^2+1)^2} = \frac{x^2+1-2x^2}{(x^2+1)^2} = \frac{1-x^2}{(x^2+1)^2}.
For x>1x>1, f(x)<0f'(x) < 0.
Thus, an=nn2+1a_n = \frac{n}{n^2+1} is decreasing for n>1n>1.
Therefore, by the Alternating Series Test, the series n=1(1)n+1nn2+1\sum_{n=1}^{\infty} (-1)^{n+1} \frac{n}{n^2+1} converges.
Since the series converges but does not converge absolutely, it converges conditionally.

3. Final Answer

The series converges conditionally.

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