39. For $n \rightarrow+\infty$ find the limits of the following functions:
1) $S_{1}(n)=\frac{1}{n}+\frac{2}{n}+\frac{3}{n}+\ldots+\frac{n-1}{n}$;
2) $S_{2}(n)=\frac{1}{n^{2}}+\frac{2}{n^{2}}+\frac{3}{n^{2}}+\ldots+\frac{n-1}{n^{2}}$;
3) $S_{3}(n)=\frac{1}{n^{3}}+\frac{2}{n^{3}}+\frac{3}{n^{3}}+\ldots+\frac{n-1}{n^{3}}$.
Solution. Each of these functions represents the sum of $n-1$ terms of an arithmetic progression. The common difference of the first progression is $\frac{1}{n}$, the second is $\frac{1}{n^{2}}$, and the third is $-\frac{1}{n^{3}}$.

By performing the addition and taking the limit, we find:

1) $S_{1}=\frac{n-1}{2}\left(\frac{1}{n}+\frac{n-1}{n}\right)=\frac{1}{2}(n-1)$;

$\lim _{n \rightarrow+\infty} S_{1}=\frac{1}{2}(\lim n-1)=+\infty$.
2) $S_{2}=\frac{n-1}{2}\left(\frac{1}{n^{2}}+\frac{n-1}{n^{2}}\right)=\frac{1}{2}\left(1-\frac{1}{n}\right)$;

$$
\lim _{n \rightarrow+\infty} S_{2}=\frac{1}{2}\left(1-\frac{1}{\lim n}\right)=\frac{1}{2}
$$

3) $S_{3}=\frac{n-1}{2}\left(\frac{1}{n^{3}}+\frac{n-1}{n^{3}}\right)=\frac{1}{2}\left(\frac{1}{n}-\frac{1}{n^{2}}\right)$;

$$
\lim _{n \rightarrow+\infty} S_{3}=\frac{1}{2}\left[\frac{1}{\operatorname{lim} n}-\frac{1}{(\lim n)^{2}}\right]=0
$$

In this problem, as $n \rightarrow+\infty$, the functions $S_{1}, S_{2}$, and $S_{3}$ are sums of infinitely small quantities, the number of which increases without bound with $n$. The results obtained show that $S_{1}$ is an infinitely large quantity, $S_{2}$ approaches $\frac{1}{2}$, and $S_{3}$ is an infinitely small quantity.

Therefore, the solution to this problem shows: if the number of addends of infinitely small quantities increases without bound, their sum can be any quantity.