Example 4.20. Investigate the convergence of the series

$$
1-\frac{3}{2}+\frac{5}{4}-\frac{7}{8}+\ldots
$$
Solution. The general term of this series can be written as

$$
(-1)^{n+1} \frac{2 n-1}{2^{n-1}},(n=1,2, \ldots)
$$

Let $a_{n}:$

Let $a_{n}=\frac{2 n-1}{2^{n-1}}$, we will prove the monotonic tendency to zero:

$$
\begin{aligned}
& a_{n}-a_{n+1}=\frac{2 n-1}{2^{n-1}}-\frac{2 n+1}{2^{n}}=\frac{4 n-2-2 n-1}{2^{n}}= \\
&=\frac{2 n-3}{2^{n}}>0(\forall n \geqslant 2)
\end{aligned}
$$

Now we will show that $\frac{n}{2^{n}} \rightarrow 0$. For this, we will show that $\forall A>0 \exists n_{0}: \forall n \geqslant n_{0} 2^{n}>A n$. Indeed, $2^{n}=(1+1)^{n}>$ $>\frac{n(n-1)}{2}$. Now it is clear that to satisfy the required inequality $2^{n}>A n$, it is sufficient that

$$
\frac{n(n-1)}{2}>A n \Rightarrow n_{0} \geqslant 2 A+1
$$

From the proven inequality, it follows that

$$
\frac{n}{2^{n}}<\frac{1}{A}=\varepsilon
$$

Since the constant $A$ can be taken arbitrarily large, then

$$
\lim _{n \rightarrow \infty} \frac{n}{2^{n}}=0
$$

In our case

$$
a_{n}=\frac{2 n-1}{2^{n-1}}=\frac{n+n-1}{2^{n-1}}=2 \frac{n}{2^{n}}+\frac{n-1}{2^{n-1}} \rightarrow 0
$$

as the sum of two infinitesimally small quantities. Thus, by Theorem 4.16, the series converges. To find the sum of the series with an absolute error $\Delta=0.001$, it is sufficient to sum

14 terms of this series, i.e., to determine $S_{14}$. This result follows from Theorem 4.16, since

$$
\left|S_{14}-S\right|<\frac{2 n-1}{2^{n}}=\frac{31}{2^{15}}<0.001
$$