Firstly I'll try to create a factorial on the bottom and rearrange to produce a sum including the series of the form $\displaystyle\sum_{n=0}^{\infty} \frac{x^n}{n!} = e^x$ . $\displaystyle\sum_{n=0}^{\infty} \frac{3^n}{(n+2)n!} = \displaystyle\sum_{n=0}^{\infty} \frac{3^n (n+1)}{(n+2)!} = \displaystyle\sum_{n=0}^{\infty} \frac{3^n (n+2-1)}{(n+2)!}$ $\ = \displaystyle\sum_{n=0}^{\infty} \frac{3^n}{(n+1)!}\ - \displaystyle\sum_{n=0}^{\infty} \frac{3^n}{(n+2)!} = \frac{1}{3} \displaystyle\sum_{n=1}^{\infty} \frac{3^n}{n!} - \frac{1}{9} \displaystyle\sum_{n=2}^{\infty} \frac{3^n}{n!}$ $\ = \frac{1}{3} \left(\displaystyle\sum_{n=0}^{\infty} \frac{3^n}{n!} -1 \right) - \frac{1}{9} \left( \displaystyle\sum_{n=0}^{\infty} \frac{3^n}{n!} - 4 \right)$ $\ = \frac{2}{9} \displaystyle\sum_{n=0}^{\infty} \frac{3^n}{n!} + \frac{1}{9} = \boxed{ \dfrac{1}{9} (2e^3 +1)}$

Why don't write a short Computer Program ? Let your computer do the job for you!

Here's the C Program :

---

include <stdio.h>

include <math.h>

int main()

{

long int g,d=1;

printf("The number of Test : ");

scanf("%ld",&g);

while(g--){

printf("The Precision (500 for highest Precision) :");

double s=.5,n=1,x;

scanf("%lf",&x);

while(x--){

double f=1,i=1;

while(i<=n){f*=i;i++;}

s+=pow(3,n)/(f*(n+2));

n++;

}

printf("%.14lf",s);

d++;

printf("\n\n");

}

return 0;

}

---

You might wanna Download the code on Pastebin.

There are thousands of ways of solving a problem!