Example 5.25. Calculate $e^{2}$ with an accuracy of 0.01.
Solution. Using expansion (2), for $x=2$ we get

$$
e^{2}=1+2+\frac{2^{2}}{2!}+\frac{2^{3}}{3!}+\ldots+\frac{2^{n}}{n!}+\ldots
$$

It remains to determine how many terms of this series need to be taken to obtain the value of $e^{2}$ with the required accuracy. Let the desired number of terms be $k$. This means that the error $\Delta S_{k}$, which we make by replacing the sum of the series with its $k$-th partial sum, is equal to the sum of the terms of the series starting from the $(k+1)$-th:

$$
\begin{gathered}
\Delta S_{k}=\frac{2^{k+1}}{(k+1)!}+\frac{2^{k+2}}{(k+2)!}+\frac{2^{k+3}}{(k+3)!}+\ldots= \\
=\frac{2^{k+1}}{(k+1)!}+\frac{2^{k+2}}{(k+1)!(k+2)}+\frac{2^{k+3}}{(k+1)!(k+2)(k+3)}+\ldots= \\
=\frac{2^{k+1}}{(k+1)!} 1+\frac{2}{k+2}+\frac{2^{2}}{(k+2)(k+3)}+\ldots .
\end{gathered}
$$

If in this series each of the numbers $k+2, k+3, \ldots$ is replaced by the number $k+1$, then the denominators of the fractions will decrease, and the fractions themselves, consequently, will increase. Hence,

$$
\Delta S_{k}100
$$

we get $k>7$.

Thus, to achieve the required accuracy, we need to take 8 terms of the series:

$$
e^{2} \approx 1+2+\frac{2^{2}}{2!}+\frac{2^{3}}{3!}+\frac{2^{4}}{4!}+\frac{2^{5}}{5!}+\frac{2^{6}}{6!}+\frac{2^{7}}{7!} \approx 7.38
$$