Example. Investigate the convergence of the series

$$
\sum_{n=1}^{\infty} \frac{1 \cdot 4 \cdot 7 \cdot \ldots \cdot(3 n-2)}{n!} \sin \frac{1}{2^{n+1}}
$$
## Solution.

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

$$
a_{n}=\frac{1 \cdot 4 \cdot 7 \cdot \ldots \cdot(3 n-2)}{n!} \sin \frac{1}{2^{n+1}}>0
$$

for all $n \geq 1$.

2. Since $\sin x \sim x$ as $x \rightarrow 0$, we can simplify the expression for $a_{n}$:

$$
\frac{1 \cdot 4 \cdot 7 \cdot \ldots \cdot(3 n-2)}{n!} \sin \frac{1}{2^{n+1}} \sim \frac{1 \cdot 4 \cdot 7 \cdot \ldots \cdot(3 n-2)}{2^{n+1} n!}
$$

i.e., we will investigate the convergence of the series $\sum_{n=1}^{\infty} b_{n}$, where

$$
b_{n}=\frac{1 \cdot 4 \cdot 7 \cdot \ldots \cdot(3 n-2)}{2^{n+1} n!}
$$

and then use the second comparison theorem.

Since $b_{n}$ contains products of factors similar to factorials, we should apply the D'Alembert's ratio test.

3. We compute $b_{n+1}$:

$$
b_{n+1}=\frac{1 \cdot 4 \cdot 7 \cdot \ldots \cdot(3 n-2) \cdot(3 n+1)}{2^{n+2}(n+1)!}
$$

4. We compute $\varrho$ using formula (1):

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

$$
\begin{aligned}
=\lim _{n \rightarrow \infty} \frac{1 \cdot 4 \cdot 7 \cdot \ldots \cdot(3 n-2) \cdot(3 n+1)}{2^{n+2}(n+1)!} \cdot \frac{2^{n+1} n!}{1 \cdot 4 \cdot 7 \cdot \ldots \cdot(3 n-2)} & \\
=\lim _{n \rightarrow \infty} \frac{3 n+1}{2(n+1)} & =\frac{3}{2}
\end{aligned}
$$

5. We apply the D'Alembert's ratio test and the second comparison theorem.

Since $\varrho=3 / 2>1$, the series

$$
\sum_{n=1}^{\infty} \frac{1 \cdot 4 \cdot 7 \cdot \ldots \cdot(3 n-2)}{2^{n+1} n!}
$$

diverges. Therefore, by the second comparison theorem, the original series also diverges.

Answer. The series $\sum_{n=1}^{\infty} \frac{1 \cdot 4 \cdot 7 \cdot \ldots \cdot(3 n-2)}{n!} \sin \frac{1}{2^{n+1}}$ diverges.

Conditions of the problems. Investigate the convergence of the series.

1. $\sum_{n=2}^{\infty} \frac{3^{n}+2}{2^{n}(n+1)!}$.
2. $\quad \sum_{n=1}^{\infty} \frac{5^{n}}{(n!)^{3}}$.
3. $\sum_{n=1}^{\infty} \frac{(3 n+2)!}{2^{n}(2 n+5)!}$.
4. $\quad \sum_{n=1}^{\infty} \frac{2^{n}\left(n^{5}-1\right)}{n!}$.
5. $\sum_{n=1}^{\infty} \frac{5^{n} n!}{(2 n)!}$.
6. $\sum_{n=1}^{\infty} \frac{n!}{3^{n}+2}$.
7. $\sum_{n=1}^{\infty} \frac{n!(2 n+2)!}{(3 n)!}$.
8. $\sum_{n=1}^{\infty} \frac{2^{n} n!}{(2 n)!}$.
9. $\sum_{n=1}^{\infty} \frac{2^{n} n!}{n^{n}}$.
10. $\sum_{n=1}^{\infty} \frac{3^{n} n!}{n^{n}}$.

Answers. 1. The series converges $(\varrho=0)$. 2. The series converges $(\varrho=0)$. 3. The series diverges $(\varrho=+\infty)$. 4. The series converges $(\varrho=0)$. 5. The series converges $(\varrho=0)$. 6. The series diverges $(\varrho=+\infty)$. 7. The series converges $(\varrho=4 / 9)$. 8. The series converges $(\varrho=0)$. 9. The series converges $(\varrho=2 / e)$. 10. The series diverges $(\varrho=3 / e)$.

## 10.5. Cauchy's Root 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 $\lim _{n \rightarrow \infty} \sqrt[n]{b_{n}}$ exists and is easily computable.

Plan of Solution. If $b_{n}$ has, for example, the form $g(n) f(n)^{n^{k}}$, where $f$ and $g$ are rational functions of $n$, then $\lim _{n \rightarrow \infty} \sqrt[n]{b_{n}}$ exists and is easily computable. In such a case, we apply the Cauchy's root test.

Let the series with positive terms be

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

If the limit

$$
\lim _{n \rightarrow \infty} \sqrt[n]{b_{n}}=\varrho
$$

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

1. We check that $a_{n}>0$ for all $n \geq 1$.
2. We 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 theorem.
3. We compute the limit

$$
\lim _{n \rightarrow \infty} \sqrt[n]{b_{n}}=\varrho
$$

4. We apply the Cauchy's root test and the second comparison theorem.

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

Remark. It is useful to keep in mind that

$$
\lim _{n \rightarrow \infty} \sqrt[n]{n}=1, \quad \lim _{n \rightarrow \infty} \sqrt[n]{\ln n}=1, \quad \lim _{n \rightarrow \infty} \sqrt[n]{P(n)}=1
$$

where $P(n)$ is a polynomial in $n$.