1. (a) (4 points) determine whether the series ∑k = 1∞ k / (k² + 2) converges or diverges.

Explanation:

Step1: Apply limit - comparison test

Let (a_{k}=\frac{k}{k^{2}+2}) and (b_{k}=\frac{1}{k}). Then (\lim_{k\rightarrow\infty}\frac{a_{k}}{b_{k}}=\lim_{k\rightarrow\infty}\frac{\frac{k}{k^{2}+2}}{\frac{1}{k}}).

Step2: Simplify the limit expression

(\lim_{k\rightarrow\infty}\frac{\frac{k}{k^{2}+2}}{\frac{1}{k}}=\lim_{k\rightarrow\infty}\frac{k\cdot k}{k^{2}+2}=\lim_{k\rightarrow\infty}\frac{k^{2}}{k^{2}+2}).

Step3: Evaluate the limit

Divide both numerator and denominator by (k^{2}): (\lim_{k\rightarrow\infty}\frac{k^{2}/k^{2}}{k^{2}/k^{2}+2/k^{2}}=\lim_{k\rightarrow\infty}\frac{1}{1 + \frac{2}{k^{2}}}=1).

Step4: Analyze the series of (b_{k})

The series (\sum_{k = 1}^{\infty}b_{k}=\sum_{k=1}^{\infty}\frac{1}{k}) is a harmonic series, which diverges.

Step5: Draw conclusion

Since (\lim_{k\rightarrow\infty}\frac{a_{k}}{b_{k}} = 1>0) and (\sum_{k = 1}^{\infty}b_{k}) diverges, by the limit - comparison test, the series (\sum_{k=1}^{\infty}\frac{k}{k^{2}+2}) diverges.

Answer:

The series (\sum_{k = 1}^{\infty}\frac{k}{k^{2}+2}) diverges.