<s>[INST] 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.
 [/INST] 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
$$=\frac{1}{2}((e^3-1)-(e-1))$$
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