1390. Write the formula for the general term of each series:
1) $1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\ldots$,
2) $\frac{2}{5}-\frac{4}{8}+\frac{6}{11}-\frac{8}{14}+\ldots$
Solution. 1) The denominators of the terms of this series are squares of natural numbers, hence the general term of the series is

$$
a_{n}=\frac{1}{n^{2}}
$$

2) The numerators of the terms of this series are even numbers of the form $2 n$, and the denominators are numbers obtained by the formula $3 n+2$ $(n=1,2,3, \ldots)$. Taking into account that the signs of the terms of the series alternate, we get

$$
a_{n}=(-1)^{n-1} \frac{2 n}{3 n+2}
$$