Example 5.19. Find the interval of convergence of the power series

$$
1+\frac{3}{1!} x+\frac{3^{2}}{2!} x^{2}+\frac{3^{3}}{3!} x^{3}+\ldots+\frac{3^{n}}{n!} x^{n}+\ldots
$$
Solution. We have $u_{n}=\frac{3^{n}}{n!}|x|^{n}, u_{n+1}=\frac{3^{n+1}}{(n+1)!}|x|^{n+1}$.

From this,

$$
l=\lim _{n \rightarrow \infty} \frac{u_{n+1}}{u_{n}}=\lim _{n \rightarrow \infty} \frac{3^{n+1} \cdot n!|x|^{n+1}}{(n+1)!3^{n}|x|^{n}}=|x| \lim _{n \rightarrow \infty} \frac{3}{n+1}=0<1
$$

Therefore, for any finite $x$, by the D'Alembert's ratio test, the given series converges absolutely. The interval of convergence of the considered series is the entire real line.

## 5.6. Expansion of Functions into Power Series of Taylor

The Taylor series for a function $f(x)$ is a power series of the form

$$
f(x)=f(0)+f^{\prime}(0) x+\frac{f^{\prime \prime}(0)}{2!} x^{2}+\frac{f^{\prime \prime \prime}(0)}{3!} x^{3}+\ldots+\frac{f^{(n)}(0)}{n!} x^{n}+\ldots
$$

To represent an elementary function as a sum of a Taylor series, the following procedure is usually followed: compute the successive derivatives of the given function at the point $x=0$, and then, using formula (1), construct the Taylor series for it and determine the interval of convergence of the obtained series. In this interval, the Taylor series converges to the function $f(x)$ that generated it, provided that all values $f(0), f^{\prime}(0), \ldots, f^{(n)}(0), \ldots$ are obtained by direct substitution of the value $x=0$ into the expressions $f(x)$, $f^{\prime}(x), \ldots, f^{(n)}(x), \ldots$

Using the described method, the Taylor series expansions for the following functions can be found:

$$
\begin{gathered}
e^{x}=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\ldots+\frac{x^{n}}{n!}+\ldots(-\infty<x<\infty) \\
\sin x=x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-\ldots+(-1)^{n} \frac{x^{2 n+1}}{(2 n+1)!}+\ldots(-\infty<x<\infty) \\
\cos x=1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-\ldots+(-1)^{n} \frac{x^{2 n}}{(2 n)!}+\ldots(-\infty<x<\infty)
\end{gathered}
$$

$$
\begin{gathered}
(1+x)^{m}=1+m x+\frac{m(m-1)}{2!} x^{2}+\ldots+ \\
+\frac{m(m-1) \ldots(m-n+1)}{n!} x^{n}+\ldots \quad(-1<x<1) \\
\frac{1}{1-x}=1+x+x^{2}+\ldots+x^{n}+\ldots \quad(-1<x<1) \\
\ln (1+x)=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\ldots+(-1)^{n-1} \frac{x^{n}}{n}+\ldots \quad(-1<x \leq 1)
\end{gathered}
$$

In addition to the method described, Taylor series expansions for functions can be obtained by starting from known expansions, such as expansions (2) - (7). For this purpose, the following operations can be performed on power series within their intervals of convergence:

1) two power series can be added term by term and multiplied according to the rules of polynomial multiplication;
2) a power series can be multiplied term by term by a common factor;
3) a power series can be integrated and differentiated term by term any number of times.

Since a power series for its sum is a Taylor series, the expansion obtained as a result of the specified operations will be the desired one.