5.1. For the sequence $\left\{a_{n}\right\}$, it is known that $a_{1}=1.5$ and $a_{n}=\frac{1}{n^{2}-1}$ for $n \in \mathbb{N}, n>1$. Are there such values of $n$ for which the sum of the first $n$ terms of this sequence differs from 2.25 by less than 0.01? If yes, find the smallest one.
Answer: yes, $n=100$.

Solution. The general formula for the terms of the sequence (except the first) can be written as $(n \geqslant 2)$:

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

As a result, the sum of the first $n$ terms of the sequence, except the first, takes the form:

$$
\begin{aligned}
& \frac{1}{2}\left[\left(1-\frac{1}{3}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{4}-\frac{1}{6}\right)+\left(\frac{1}{5}-\frac{1}{7}\right)+\right. \\
& \left.\quad+\ldots+\left(\frac{1}{n-3}-\frac{1}{n-1}\right)+\left(\frac{1}{n-2}-\frac{1}{n}\right)+\left(\frac{1}{n-1}-\frac{1}{n+1}\right)\right]
\end{aligned}
$$

After simplifications, for the sum of the first $n$ terms of the sequence, we can write:

$$
S_{n}=1.5+\frac{1}{2}\left[1+\frac{1}{2}-\frac{1}{n}-\frac{1}{n+1}\right]=2.25-\frac{1}{2}\left(\frac{1}{n}+\frac{1}{n+1}\right)
$$

Let $f(n)=\frac{1}{2}\left(\frac{1}{n}+\frac{1}{n+1}\right)$. Then since $f(n)$ is decreasing and

$$
\begin{aligned}
f(100) & =\frac{1}{2}\left(\frac{1}{100}+\frac{1}{101}\right)\frac{1}{2}\left(\frac{1}{100}+\frac{1}{100}\right)=\frac{1}{100}
\end{aligned}
$$

the desired value of $n$ is 100.