Question

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) $\sum\limits_{n=1}^{\infty} \frac{2n^3}{\sqrt{3n + n^7}}$

show all your work clearly.

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) _____________________________
(c) _____________________________

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 \(p\leqslant1\). 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.

Part (b)

Step 1: Make a substitution

Let \(m=n - 1\). When \(n = 2\), \(m = 1\). So the series \(\sum_{n=2}^{\infty}\frac{1}{(n - 1)^4}\) can be rewritten as \(\sum_{m = 1}^{\infty}\frac{1}{m^4}\)

Step 2: Apply the p - series test

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

Part (c)

Step 1: Analyze the leading terms

For large \(n\), the dominant terms in the numerator and denominator are considered. The numerator is \(2n^3\) and the denominator is \(\sqrt{n^7}=\ n^{7/2}\) (since for large \(n\), \(3n + n^7\approx n^7\) and \(\sqrt{3n + n^7}\approx\sqrt{n^7}=n^{7/2}\))

Step 2: Find the limit of the ratio of \(a_n\) to a simpler term

We consider the limit \(\lim_{n
ightarrow\infty}\frac{\frac{2n^3}{\sqrt{3n + n^7}}}{\frac{1}{n^{1/2}}}=\lim_{n
ightarrow\infty}\frac{2n^3\cdot n^{1/2}}{\sqrt{n^7}}=\lim_{n
ightarrow\infty}\frac{2n^{7/2}}{n^{7/2}} = 2\)

Step 3: Apply the limit comparison test

We compare with the series \(\sum_{n = 1}^{\infty}\frac{1}{n^{1/2}}\). The series \(\sum_{n=1}^{\infty}\frac{1}{n^{1/2}}\) is a p - series with \(p=\frac{1}{2}\leqslant1\), so it diverges. Since \(\lim_{n
ightarrow\infty}\frac{a_n}{\frac{1}{n^{1/2}}}=2>0\) (a finite positive number), by the limit comparison test, the series \(\sum_{n = 1}^{\infty}\frac{2n^3}{\sqrt{3n + n^7}}\) diverges.

Answer:

s:
(a) The series converges (by p - series test with \(p=\frac{3}{2}>1\))
(b) The series converges (by p - series test with \(p = 4>1\) after substitution)
(c) The series diverges (by limit comparison test with \(\sum\frac{1}{n^{1/2}}\))