TITLE: Sum the series: $1+\frac{1+3}{2!}+\frac{1+3+3^2}{3!}+\cdots$
QUESTION [5 upvotes]: 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.

REPLY [2 votes]: The general term is
$$T_n={1+3+\cdots +3^n\over n!}={3^{n+1}-1\over 2}{1\over (n+1)!}$$
Now we can rewrite the series as
$$S_n={1\over 2}\sum_{n=0}^\infty {3^{n+1}\over (n+1)!}-{1\over 2}\sum_{n=0}^\infty {1\over (n+1!}={1\over 2}\left(e^3-1\right)-{1\over 2}(e-1)={1\over 2}\left(e^3-e\right)$$