Example. Find the sum of the series

$$
\sum_{n=1}^{\infty} \frac{72}{n^{2}+5 n+4}
$$
Solution.

1. The roots of the denominator $n=-1$ and $n=-4$ differ by an integer, i.e., $n^{2}+5 n+4=(n+1)(n+1+3)$. Therefore, the terms of the sequence of partial sums of the series $\sum_{n=1}^{\infty} a_{n}$ are easily found, as many terms in the expression $S_{n}=a_{1}+a_{2}+\ldots+a_{n}$ cancel each other out.
2. We decompose the general term of the series into elementary fractions

$$
a_{n}=\frac{72}{n^{2}+5 n+4}=\frac{24}{n+1}-\frac{24}{n+4}
$$

and list several terms of the series:

$$
a_{1}=\frac{24}{2}-\frac{24}{5}, \quad a_{2}=\frac{24}{3}-\frac{24}{6}, \quad a_{3}=\frac{24}{4}-\frac{24}{7}
$$

$$
\begin{aligned}
& a_{4}=\frac{24}{5}-\frac{24}{8}, \quad a_{5}=\frac{24}{6}-\frac{24}{9}, \quad a_{6}=\frac{24}{7}-\frac{24}{10} \\
& \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \\
& a_{n-5}=\frac{24}{n-4}-\frac{24}{n-1}, \quad a_{n-4}=\frac{24}{n-3}-\frac{24}{n} \\
& a_{n-3}=\frac{24}{n-2}-\frac{24}{n+1}, \quad a_{n-2}=\frac{24}{n-1}-\frac{24}{n+2} \\
& a_{n-1}=\frac{24}{n}-\frac{24}{n+3}, \quad a_{n}=\frac{24}{n+1}-\frac{24}{n+4}
\end{aligned}
$$

3. By canceling all possible terms, we find the $n$-th partial sum of the series:

$S_{n}=a_{1}+a_{2}+\ldots+a_{n}=$

$=\frac{24}{2}+\frac{24}{3}+\frac{24}{4}-\frac{24}{n+2}-\frac{24}{n+3}-\frac{24}{n+4}=26-\frac{24}{n+2}-\frac{24}{n+3}-\frac{24}{n+4}$.

4. We compute the sum of the series using formula (1):

$$
S=\lim _{n \rightarrow \infty} S_{n}=\lim _{n \rightarrow \infty}\left(26-\frac{24}{n+2}-\frac{24}{n+3}-\frac{24}{n+4}\right)=26
$$

Answer. $S=26$.

Problem Statements. Find the sums of the series.

1. $\sum_{n=1}^{\infty} \frac{6}{n^{2}+5 n+6}$.
2. $\quad \sum_{n=6}^{\infty} \frac{3}{n^{2}-5 n+6}$
3. $\quad \sum_{n=1}^{\infty} \frac{30}{25 n^{2}+5 n-6}$.
4. $\quad \sum_{n=1}^{\infty} \frac{4}{4 n^{2}-1}$.
5. $\quad \sum_{n=1}^{\infty} \frac{18}{n^{2}+3 n}$.
6. $\quad \sum_{n=1}^{\infty} \frac{90}{4 n^{2}+8 n-5}$.
7. $\quad \sum_{n=1}^{\infty} \frac{3}{9 n^{2}-3 n-2}$.
8. $\sum_{n=1}^{\infty} \frac{16}{16 n^{2}-8 n-3}$.
9. $\sum_{n=1}^{\infty} \frac{8}{n(n+1)(n+2)}$.
10. $\sum_{n=1}^{\infty} \frac{60}{(2 n+1)(2 n+3)(2 n+5)}$.

Answers. 1. $S=2$. 2. $S=1$. 3. $S=2$. 4. $S=2$. 5. $S=11$. 6. $S=23$. 7. $S=1$. 8. $S=4$. 9. $S=2$. 10. $S=1$.

## 10.2. First Comparison Theorem

Problem Statement. Investigate the convergence of the series with non-negative terms

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

where $a_{n}=f\left(n, u_{1}(n), u_{2}(n), \ldots\right)$ and $u_{1}(n), u_{2}(n), \ldots-$ are functions with known minimum and maximum values (e.g., sine, cosine, etc.), and the function $f$ monotonically depends on $u_{1}, u_{2}, \ldots$

## Plan of Solution.

1. Check that $\lim _{n \rightarrow \infty} a_{n}=0$ (if $\lim _{n \rightarrow \infty} a_{n} \neq 0$, then the series diverges, as the necessary condition for the convergence of the series is not satisfied).
2. Since $a_{n} \geq 0$, apply the first comparison theorem.

Let there be two series with non-negative terms $\sum_{n=1}^{\infty} a_{n} \quad u$ $\sum_{n=1}^{\infty} b_{n}$

If $a_{n} \leq b_{n}$, then the convergence of the series $\sum_{n=1}^{\infty} b_{n}$ implies the convergence of the series $\sum_{n=1}^{\infty} a_{n}$.

If $a_{n} \geq b_{n}$, then the divergence of the series $\sum_{n=1}^{\infty} b_{n}$ implies the divergence of the series $\sum_{n=1}^{\infty} a_{n}$.

3. To draw a conclusion about the convergence (divergence) of the given series, we must establish the validity of one of the two hypotheses (we check them in any order).

I. The given series $\sum_{n=1}^{\infty} a_{n}$ converges.

II. The given series $\sum_{n=1}^{\infty} a_{n}$ diverges.

I. We check the first hypothesis. To establish that the given series $\sum_{n=1}^{\infty} a_{n}$ converges, we need to find a convergent series $\sum_{n=1}^{\infty} b_{n}$ such that

$$
a_{n} \leq b_{n}
$$

As the reference series $\sum_{n=1}^{\infty} b_{n}$, we use one of the following series:

a) convergent harmonic series $\sum_{n=1}^{\infty} \frac{c}{n^{p}}$ for $p>1$;
b) convergent geometric series $\sum_{n=1}^{\infty} c q^{n}$ for $0<q<1$;
c) convergent series $\sum_{n=1}^{\infty} \frac{c}{n(n+1)}$ (converges to $c$);
d) convergent series $\sum_{n=1}^{\infty} \frac{c}{n(n+1)(n+2)}$ (converges to $\frac{c}{2}$).

II. We check the second hypothesis. To establish that the given series $\sum_{n=1}^{\infty} a_{n}$ diverges, we need to find a divergent series $\sum_{n=1}^{\infty} b_{n}$ such that

$$
a_{n} \geq b_{n}
$$

As the reference series $\sum_{n=1}^{\infty} b_{n}$, we use one of the following series:

a) divergent harmonic series $\sum_{n=1}^{\infty} \frac{c}{n}$;
b) divergent series $\sum_{n=1}^{\infty} \frac{c}{n(n+1)}$ (diverges to $\infty$);
c) divergent series $\sum_{n=1}^{\infty} \frac{c}{n(n+1)(n+2)}$ (diverges to $\infty$).

4. To find the reference series, we use the known minimum and maximum values of the functions $u_{1}(n), u_{2}(n), \ldots$:

$$
\begin{aligned}
& 0 \leq \sin n \leq 1, \quad 0 \leq \cos n \leq 1, \quad 0 \leq \sin ^{2} n \leq 1, \quad 0 \leq \cos ^{2} n \leq 1, \\
& 0 \leq \sin ^{2} n+\cos ^{2} n=1, \quad 0 \leq \sin ^{2} n-\cos ^{2} n \leq 1, \quad 0 \leq \sin ^{2} n \cos ^{2} n \leq \frac{1}{4}, \\
& 0 \leq \sin ^{2} n \cos ^{2} n \leq \frac{1}{4}, \quad 0 \leq \sin ^{2} n \cos ^{2} n \leq \frac{1}{4}, \quad 0 \leq \sin ^{2} n \cos ^{2} n \leq \frac{1}{4}, \\
& 0 \leq \sin ^{2} n \cos ^{2} n \leq \frac{1}{4}, \quad 0 \leq \sin ^{2} n \cos ^{2} n \leq \frac{1}{4}, \quad 0 \leq \sin ^{2} n \cos ^{2} n \leq \frac{1}{4}, \\
& 0 \leq \sin ^{2} n \cos ^{2} n \leq \frac{1}{4}, \quad 0 \leq \sin ^{2} n \cos ^{2} n \leq \frac{1}{4}, \quad 0 \leq \sin ^{2} n \cos ^{2} n \leq \frac{1}{4}, \\
& 0 \leq \sin ^{2} n \cos ^{2} n \leq \frac{1}{4}, \quad 0 \leq \sin ^{2} n \cos ^{2} n \leq \frac{1}{4}, \quad 0 \leq \sin ^{2} n \cos ^{2} n \leq \frac{1}{4}, \\
& 0 \leq \sin ^{2} n \cos ^{2} n \leq \frac{1}{4}, \quad 0 \leq \sin ^{2} n \cos ^{2} n \leq \frac{1}{4}, \quad 0 \leq \sin ^{2} n \cos ^{2} n \leq \frac{1}{4}, \\
& 0 \leq \sin ^{2} n \cos ^{2} n \leq \frac{1}{4}, \quad 0 \leq \sin ^{2} n \cos ^{2} n \leq \frac{1}{4}, \quad 0 \leq \sin ^{2} n \cos ^{2} n \leq \frac{1}{4}, \\
& 0 \leq \sin ^{2} n \cos ^{2} n \leq \frac{1}{4}, \quad 0 \leq \sin ^{2} n \cos ^{2} n \leq \frac{1}{4}, \quad 0 \leq \sin ^{2} n \cos ^{2} n \leq \frac{1}{4}, \\
& 0 \leq \sin