two similar - looking series are given. test each one for convergence or divergence.\n(a) $sum_{n =…

Explanation:

Step1: Use limit - comparison test for (a)

For the series $\sum_{n = 1}^{\infty}\frac{n}{n^{2}+3}$, compare it with the series $\sum_{n = 1}^{\infty}\frac{1}{n}$. Calculate the limit $\lim_{n\rightarrow\infty}\frac{\frac{n}{n^{2}+3}}{\frac{1}{n}}=\lim_{n\rightarrow\infty}\frac{n^{2}}{n^{2}+3}$.

Step2: Evaluate the limit for (a)

$\lim_{n\rightarrow\infty}\frac{n^{2}}{n^{2}+3}=\lim_{n\rightarrow\infty}\frac{1}{1 + \frac{3}{n^{2}}}=1$. Since $\sum_{n = 1}^{\infty}\frac{1}{n}$ is a harmonic series (divergent) and the limit is a positive - finite number, by the limit - comparison test, $\sum_{n = 1}^{\infty}\frac{n}{n^{2}+3}$ is divergent.

Step3: Use the root - test for (b)

For the series $\sum_{n = 1}^{\infty}(\frac{n}{n^{2}+3})^{n}$, apply the root - test. Calculate $\lim_{n\rightarrow\infty}\sqrt[n]{a_{n}}=\lim_{n\rightarrow\infty}\frac{n}{n^{2}+3}$.

Step4: Evaluate the limit for (b)

$\lim_{n\rightarrow\infty}\frac{n}{n^{2}+3}=\lim_{n\rightarrow\infty}\frac{\frac{1}{n}}{1+\frac{3}{n^{2}}}=0$. Since $\lim_{n\rightarrow\infty}\sqrt[n]{a_{n}} = 0<1$, by the root - test, $\sum_{n = 1}^{\infty}(\frac{n}{n^{2}+3})^{n}$ is convergent.

Answer:

(a) divergent (b) convergent