show the power series converges for |x|<1, and find the closed forms

Question

\begin{matrix} 1) S(x) = \sum_{n=0}^\infty (1+ \frac{1}{1!} + \frac{1}{2!} + ...+ \frac{1}{n!}) x^n \\ 2) S(x) =\sum_{n=1}^\infty a_n x^n, a_n= \Big\{ \begin{alignedat}{3} -2/n ,n=3k \\ 1/n,n \neq 3k \end{alignedat}\Big\} \end{matrix} I know that for the first one, the idea of cauchy product is needed, and for the second, I should probably write the series as a combination of other series. But I am still confused and would appreciate some help.

Answer 1

answer ad question 1)

After noting that $\sum_{n=0}^\infty \frac{1}{n!} x^n = e^x $ for all $x\in\mathbb{R}$ let us consider

$$\sum_{n=1}^\infty \frac{1}{(n-1)!} x^n = \sum_{n=1}^\infty \frac{1}{(n-1)!} x^n = x \sum_{m=0}^\infty \frac{1}{m!} x^{m} = x e^x$$

and

\begin{eqnarray} \sum_{n=2}^\infty \frac{1}{(n-2)!} x^n = x^2 \sum_{n=2}^\infty \frac{1}{(n-2)!} x^{n-2} = x^2 \sum_{m=0}^\infty \frac{1}{m!} x^{m} = x^2 e^x \end{eqnarray}

and in general

\begin{eqnarray} \sum_{n=k}^\infty \frac{1}{(n-k)!} x^n = x^k \sum_{n=k}^\infty \frac{1}{(n-k)!} x^{n-k} = x^k \sum_{m=0}^\infty \frac{1}{m!} x^{m} = x^k e^x \end{eqnarray}

As all of the individual series are uniformly convergent for all $x$, we obtain \begin{eqnarray} S(x) &=& \sum_{n=0}^\infty (1+ \frac{1}{1!} + \frac{1}{2!} + ...+ \frac{1}{n!}) x^n \\ &=& e^x \sum_{k=0}^\infty x^k = e^x \frac{1}{1-x} \end{eqnarray} where the last equality holds if and only if the geometric series with terms $x^k$ converges, which it does for $|x|<1$.