real analysis - Sum the series: $1+frac{1+3}{2!}+frac{1+3+3^2}{3!}+cdots$

We have the series $$1+\frac{1+3}{2!}+\frac{1+3+3^2}{3!}+\cdots$$ How can we find the sum$?$

MY TRY: $n$th term of the series i.e $T_n=\frac{3^0+3^1+3^2+\cdots+3^n}{(n+1)!}$. I don't know how to proceed further. Thank you.

Answer

The numerator is a geometric sum that evaluates to,

$$\frac{3^{n+1}-1}{3-1}$$

Hence what we have is,

$$\frac{1}{2} \sum_{n=0}^{\infty} \frac{(3^{n+1}-1)}{(n+1)!}$$

$$=\frac{1}{2} \sum_{n=1}^{\infty} \frac{3^n-1}{n!}$$

$$=\frac{1}{2} \left( \sum_{n=1}^{\infty} \frac{3^n}{n!}- \sum_{n=1}^{\infty} \frac{1^n}{n!} \right)$$

Recognizing the Taylor series of $e^x$ we have

$$=\frac{1}{2}((e^3-1)-(e-1))$$