Question

18 pts
determine whether the series converges or diverges. state the test you used. show all your work and explain your reasons clearly.
(a) $1 + \frac{1}{2\sqrt{2}} + \frac{1}{3\sqrt{3}} + \frac{1}{4\sqrt{4}} + \frac{1}{5\sqrt{5}} \dots$

(b) $\sum\limits_{n=2}^{\infty} \frac{1}{(n - 1)^4}$

(c) $\sum\limits_{n=1}^{\infty} \frac{2n^2}{\sqrt{3n + n^3}}$

show all your work clearly.
\\(\implies\\)you must show all your work on your paper for part (a)-(c). then you need to enter your answers for part (a) and (c) whether each series converges or diverges:
(a)
(b)

Explanation:

Part (a)

Step 1: Identify the general term

The series is \(1+\frac{1}{2\sqrt{2}}+\frac{1}{3\sqrt{3}}+\frac{1}{4\sqrt{4}}+\frac{1}{5\sqrt{5}}+\dots\). The general term \(a_n=\frac{1}{n\sqrt{n}}=\frac{1}{n^{3/2}}\) for \(n = 1,2,3,\dots\).

Step 2: Apply the p - series test

The p - series test states that the series \(\sum_{n = 1}^{\infty}\frac{1}{n^p}\) converges if \(p>1\) and diverges if \(p\leq1\). For our series, \(p=\frac{3}{2}\). Since \(\frac{3}{2}>1\), by the p - series test, the series \(\sum_{n = 1}^{\infty}\frac{1}{n^{3/2}}\) converges.

Step 1: Make a substitution

Let \(k=n - 1\). When \(n = 2\), \(k=1\); as \(n
ightarrow\infty\), \(k
ightarrow\infty\). So the series \(\sum_{n = 2}^{\infty}\frac{1}{(n - 1)^4}\) becomes \(\sum_{k = 1}^{\infty}\frac{1}{k^4}\).

Step 2: Apply the p - series test

For the p - series \(\sum_{k = 1}^{\infty}\frac{1}{k^p}\), here \(p = 4\). Since \(p=4>1\), by the p - series test, the series \(\sum_{k = 1}^{\infty}\frac{1}{k^4}\) (and thus the original series \(\sum_{n = 2}^{\infty}\frac{1}{(n - 1)^4}\)) converges.

Step 1: Analyze the leading terms

For large \(n\), the dominant terms in the numerator and denominator are considered. The numerator is \(2n^2\) and the denominator is \(\sqrt{n^3}=\ n^{3/2}\) (since for large \(n\), \(n^3\) dominates \(3n\) in \(\sqrt{3n + n^3}\)). So we can use the limit comparison test. Let \(a_n=\frac{2n^2}{\sqrt{3n + n^3}}\) and \(b_n=\frac{2n^2}{n^{3/2}}=2n^{2-\frac{3}{2}}=2n^{\frac{1}{2}}\).

Step 2: Calculate the limit of \(\frac{a_n}{b_n}\)

\(\lim_{n
ightarrow\infty}\frac{a_n}{b_n}=\lim_{n
ightarrow\infty}\frac{\frac{2n^2}{\sqrt{3n + n^3}}}{2n^{\frac{1}{2}}}=\lim_{n
ightarrow\infty}\frac{n^2}{n^{\frac{1}{2}}\sqrt{n^3(1+\frac{3}{n^2})}}=\lim_{n
ightarrow\infty}\frac{n^2}{n^{\frac{1}{2}}\cdot n^{\frac{3}{2}}\sqrt{1+\frac{3}{n^2}}}=\lim_{n
ightarrow\infty}\frac{n^2}{n^2\sqrt{1+\frac{3}{n^2}}}=\lim_{n
ightarrow\infty}\frac{1}{\sqrt{1+\frac{3}{n^2}}}=1\)

Step 3: Analyze the convergence of \(\sum b_n\)

The series \(\sum_{n = 1}^{\infty}b_n=\sum_{n = 1}^{\infty}2n^{\frac{1}{2}}\) is a p - series with \(p=\frac{1}{2}\leq1\). So \(\sum b_n\) diverges.

Step 4: Apply the limit comparison test

Since \(\lim_{n
ightarrow\infty}\frac{a_n}{b_n}=1>0\) and \(\sum b_n\) diverges, by the limit comparison test, \(\sum a_n\) (the original series) diverges.

Answer:

The series converges (by the p - series test with \(p=\frac{3}{2}>1\))

Part (b)