(16 points) determine whether the series is absolutely converging, conditionally converging, or…

(a)

Explanation:

Step1: Check for absolute convergence

Consider the series of absolute values (\sum_{n = 1}^{\infty}\left|\frac{(- 1)^{n}n\ln n}{n^{2}+1}\right|=\sum_{n = 1}^{\infty}\frac{n\ln n}{n^{2}+1}). Use the limit - comparison test. Compare with the series (\sum_{n = 2}^{\infty}\frac{1}{n}) (since for (n\geq2), (\ln n>0)). Calculate (\lim_{n\rightarrow\infty}\frac{\frac{n\ln n}{n^{2}+1}}{\frac{1}{n}}=\lim_{n\rightarrow\infty}\frac{n^{2}\ln n}{n^{2}+1}). Divide numerator and denominator by (n^{2}): (\lim_{n\rightarrow\infty}\frac{\ln n}{1 + \frac{1}{n^{2}}}=\infty) (because (\lim_{n\rightarrow\infty}\ln n=\infty)). Since (\sum_{n = 2}^{\infty}\frac{1}{n}) is a divergent (p -)series ((p = 1)), the series of absolute values (\sum_{n = 1}^{\infty}\frac{n\ln n}{n^{2}+1}) diverges.

Step2: Check for conditional convergence (using the Alternating - Series Test)

Let (a_{n}=\frac{n\ln n}{n^{2}+1}). First, find (\lim_{n\rightarrow\infty}a_{n}). (\lim_{n\rightarrow\infty}\frac{n\ln n}{n^{2}+1}=\lim_{n\rightarrow\infty}\frac{\ln n}{n+\frac{1}{n}}) (using L'Hopital's rule: (\lim_{x\rightarrow\infty}\frac{\ln x}{x+\frac{1}{x}}=\lim_{x\rightarrow\infty}\frac{\frac{1}{x}}{1-\frac{1}{x^{2}}}=0)). Next, check if (a_{n + 1}<a_{n}) for (n) large enough. (a_{n}=\frac{n\ln n}{n^{2}+1}), (a_{n + 1}=\frac{(n + 1)\ln(n + 1)}{(n + 1)^{2}+1}) (a_{n}-a_{n + 1}=\frac{n\ln n}{n^{2}+1}-\frac{(n + 1)\ln(n + 1)}{(n + 1)^{2}+1}=\frac{n\ln n\left[(n + 1)^{2}+1\right]-(n + 1)\ln(n + 1)(n^{2}+1)}{(n^{2}+1)\left[(n + 1)^{2}+1\right]}) For (n) large, consider the function (f(x)=\frac{x\ln x}{x^{2}+1}), (f^{\prime}(x)=\frac{(\ln x + 1)(x^{2}+1)-2x^{2}\ln x}{(x^{2}+1)^{2}}=\frac{x^{2}+1+(x^{2}+1)\ln x-2x^{2}\ln x}{(x^{2}+1)^{2}}=\frac{x^{2}+1-(x^{2}-1)\ln x}{(x^{2}+1)^{2}}) For (x) large, ((x^{2}-1)\ln x>x^{2}+1), so (f^{\prime}(x)<0) (the function (y = f(x)) is decreasing for (x) large). By the Alternating - Series Test, (\sum_{n = 1}^{\infty}\frac{(-1)^{n}n\ln n}{n^{2}+1}) converges.

Answer:

The series (\sum_{n = 1}^{\infty}\frac{(-1)^{n}n\ln n}{n^{2}+1}) is conditionally convergent.

(b)

Explanation:

Step1: Check for absolute convergence

Consider the series of absolute values (\sum_{n = 1}^{\infty}\left|\frac{(-1)^{n}2^{n}n!}{(2n)!}\right|=\sum_{n = 1}^{\infty}\frac{2^{n}n!}{(2n)!}) Use the Ratio Test. Let (a_{n}=\frac{2^{n}n!}{(2n)!}) (a_{n+1}=\frac{2^{n + 1}(n + 1)!}{(2(n + 1))!}=\frac{2^{n+1}(n + 1)!}{(2n+2)!}) Calculate (\lim_{n\rightarrow\infty}\left|\frac{a_{n + 1}}{a_{n}}\right|=\lim_{n\rightarrow\infty}\frac{\frac{2^{n+1}(n + 1)!}{(2n+2)!}}{\frac{2^{n}n!}{(2n)!}}) Since ((2n + 2)!=(2n+2)(2n + 1)(2n)!) and ((n + 1)!=(n + 1)n!) (\lim_{n\rightarrow\infty}\left|\frac{a_{n + 1}}{a_{n}}\right|=\lim_{n\rightarrow\infty}\frac{2^{n+1}(n + 1)!}{(2n+2)!}\cdot\frac{(2n)!}{2^{n}n!}=\lim_{n\rightarrow\infty}\frac{2(n + 1)}{(2n+2)(2n + 1)}=\lim_{n\rightarrow\infty}\frac{2(n + 1)}{2(n + 1)(2n + 1)}=\lim_{n\rightarrow\infty}\frac{1}{2n+1}=0)

Answer:

Since (\lim_{n\rightarrow\infty}\left|\frac{a_{n + 1}}{a_{n}}\right| = 0<1), the series (\sum_{n = 1}^{\infty}\frac{(-1)^{n}2^{n}n!}{(2n)!}) is absolutely convergent.