1389. Write the first five terms of the series $\sum_{n=1}^{\infty}(-1)^{n-1} \frac{n}{n+1}$.
Solution. The general term $a_{n}=(-1)^{n-1} \frac{n}{n+1}$. Setting $n=1$, we get $a_{1}=(-1)^{0} \frac{1}{1+1}=\frac{1}{2}$. If $n=2$, then $a_{2}=$
$=(-1) \frac{2}{2+1}=-\frac{2}{3} \quad$ and $\quad$ further $\quad$ (for $\quad n=3 ; 4 ; 5$ ) $a_{3}=$
$=(-1)^{2} \frac{3}{3+1}=\frac{3}{4}, \quad a_{4}=(-1)^{3} \frac{4}{4+1}=-\frac{4}{5}, \quad a_{5}=$
$=(-1)^{4} \frac{5}{5+1}=\frac{5}{6}$. Therefore,

$$
\sum_{n=1}^{\infty}(-1)^{n-1} \frac{n}{n+1}=\frac{1}{2}-\frac{2}{3}+\frac{3}{4}-\frac{4}{5}+\frac{5}{6}-\ldots
$$