Example. Investigate the convergence of the series

$$
\sum_{n=1}^{\infty} \arcsin \frac{n}{\left(n^{2}+3\right)^{5 / 2}}
$$
SOLUTION.

1. We have

$$
\lim _{n \rightarrow \infty} \arcsin \frac{n}{\left(n^{2}+3\right)^{5 / 2}}=0
$$

2. We check that the terms of the given series are positive. Indeed,

$$
\arcsin \frac{n}{\left(n^{2}+3\right)^{5 / 2}}>0
$$

for all $n \geq 1$, since $n /\left(n^{2}+3\right)^{5 / 2} \in(0,1)$.

3. We conclude about the convergence or divergence of the given series using the second (limit) comparison test.

We have

$$
\arcsin \frac{n}{\left(n^{2}+3\right)^{5 / 2}} \sim \frac{1}{n^{4}} \text { as } n \rightarrow \infty
$$

The series

$$
\sum_{n=1}^{\infty} \frac{1}{n^{4}}
$$

converges as a harmonic series with $p=4>1$. Therefore, by the second (limit) comparison test, the original series also converges.

Answer. The series $\sum_{n=1}^{\infty} \arcsin \frac{n}{\left(n^{2}+3\right)^{5 / 2}}$ converges.

Conditions of the Problems. Investigate the convergence of the series.

1. $\sum_{n=1}^{\infty} \ln \frac{n^{3}+6}{n^{3}+5}$.
2. $\quad \sum_{n=1}^{\infty} n\left(1-\cos \frac{\pi}{n^{2}}\right)$.
3. $\quad \sum_{n=1}^{\infty} \sqrt[3]{n^{2}} \operatorname{arctg} \frac{1}{n^{4}}$.
4. $\quad \sum_{n=1}^{\infty} n^{2} \sin \frac{1}{\sqrt[3]{n^{7}}}$.
5. $\quad \sum_{n=1}^{\infty} n \ln \frac{n^{4}+3}{n^{4}+2}$.
6. $\quad \sum_{n=1}^{\infty} \sin \frac{n+3}{n(n+2)^{3}}$.
7. $\quad \sum_{n=1}^{\infty} \frac{3^{n}+n}{7^{n}+2 n}$.
8. $\sum_{n=1}^{\infty} \sqrt[3]{n} \arcsin \frac{n^{2}+1}{n^{4}-2}$.
9. $\quad \sum_{n=1}^{\infty} n^{2}\left(e^{1 / 2 n^{3}}-1\right)$.
10. $\sum_{n=1}^{\infty} n \operatorname{tg}^{2} \frac{\pi}{n^{2}}$.

Answers. 1. The series converges. 2. The series converges. 3. The series converges. 4. The series diverges. 5. The series converges. 6. The series converges. 7. The series converges. 8. The series converges. 9. The series diverges. 10. The series converges.

## 10.4. D'Alembert's Ratio Test

PROBLEM STATEMENT. Investigate the convergence of the series with positive terms

$$
\sum_{n=1}^{\infty} a_{n}
$$

where $a_{n} \sim b_{n}$ and $b_{n}$ contains products of many factors (e.g., factorials).

PLAN OF SOLUTION. If, when computing the limit

$$
\lim _{n \rightarrow \infty} \frac{b_{n+1}}{b_{n}}
$$

many factors in the numerator and denominator of the fraction $b_{n+1} / b_{n}$ can be canceled, then D'Alembert's ratio test is usually applied.

Let the given series with positive terms

$$
\sum_{n=1}^{\infty} b_{n}
$$

If the limit

$$
\lim _{n \rightarrow \infty} \frac{b_{n+1}}{b_{n}}=\varrho
$$

exists, then the series converges if $\varrho<1$ and diverges if $\varrho>1$. (If $\varrho=1$, D'Alembert's ratio test is inconclusive.)

1. Check that $a_{n}>0$ for all $n \geq 1$.
2. Simplify, if necessary, the expression for $a_{n}$, i.e., we will investigate the convergence of the series $\sum_{n=1}^{\infty} b_{n}$ such that $a_{n} \sim b_{n}$ as $n \rightarrow \infty$, and then use the second comparison test.
3. Compute $b_{n+1}$.
4. Compute the limit

$$
\lim _{n \rightarrow \infty} \frac{b_{n+1}}{b_{n}}=\varrho
$$

5. Apply D'Alembert's ratio test and the second comparison test.

If $\varrho<1$, the series $\sum_{n=1}^{\infty} b_{n}$ converges. Therefore, by the second comparison test, the original series $\sum_{n=1}^{\infty} a_{n}$ also converges.

If $\varrho>1$, the series $\sum_{n=1}^{\infty} b_{n}$ diverges. Therefore, by the second comparison test, the original series $\sum_{n=1}^{\infty} a_{n}$ also diverges.