Solution of $\sum_{x=2}^\infty\frac{3^x}{(x-2)!}$

Question

I would like to solve the series $\sum_{x=2}^\infty\frac{3^x}{(x-2)!}$. I know that it converges by the ratio test, but how can I find the solution of the series?

Answer 1

Set $x-2=y$

$$ \sum_{x=2}^\infty\frac{3^x}{(x-2)!}=3^2\cdot\sum_{y=0}^\infty\dfrac{3^y}{y!}$$

Finally $$e^z=\sum_{r=0}^\infty\dfrac{z^r}{r!}$$

Answer 2

Hint: Use the fact that$$e^z=\sum_{i=0}^\infty\frac{z^i}{i!}$$