(b) (5 points) show that the series ∑_{k = 1}^∞ (-1)^k * k / (k^2 + 2) converges.

Explanation:

Step1: Check the conditions of the Alternating - Series Test

Let $a_k=\frac{k}{k^{2}+2}$.

Step2: Show $\lim_{k\rightarrow\infty}a_k = 0$

We have $\lim_{k\rightarrow\infty}a_k=\lim_{k\rightarrow\infty}\frac{k}{k^{2}+2}=\lim_{k\rightarrow\infty}\frac{\frac{k}{k^{2}}}{\frac{k^{2}}{k^{2}}+\frac{2}{k^{2}}}=\lim_{k\rightarrow\infty}\frac{\frac{1}{k}}{1 + \frac{2}{k^{2}}}=0$.

Step3: Show $a_{k + 1}\leq a_k$ for $k\geq N$ (some positive integer $N$)

Find the derivative of $f(x)=\frac{x}{x^{2}+2}$ using the quotient - rule. If $f(x)=\frac{u}{v}$ where $u = x$ and $v=x^{2}+2$, then $f^\prime(x)=\frac{u^\prime v - uv^\prime}{v^{2}}=\frac{1\cdot(x^{2}+2)-x\cdot(2x)}{(x^{2}+2)^{2}}=\frac{x^{2}+2 - 2x^{2}}{(x^{2}+2)^{2}}=\frac{2 - x^{2}}{(x^{2}+2)^{2}}$. The critical points of $f(x)$ are found by setting $f^\prime(x)=0$, so $2 - x^{2}=0$, which gives $x=\sqrt{2}$ (we consider the positive value since $k\in\mathbb{N}$). For $x\geq2$ (since $k\in\mathbb{N}$), $f^\prime(x)<0$, so the function $y = f(x)$ is decreasing. Thus, $a_{k+1}=\frac{k + 1}{(k + 1)^{2}+2}\leq a_k=\frac{k}{k^{2}+2}$ for $k\geq2$. Since the two conditions of the Alternating - Series Test are satisfied (i.e., $\lim_{k\rightarrow\infty}a_k = 0$ and $a_{k+1}\leq a_k$ for $k$ large enough), the series $\sum_{k = 1}^{\infty}(-1)^{k}\frac{k}{k^{2}+2}$ converges.

Answer:

The series $\sum_{k = 1}^{\infty}(-1)^{k}\frac{k}{k^{2}+2}$ converges by the Alternating - Series Test.