<s>[INST] Sum of increasing geometric series Sum of the series 
$1 + \frac{1+3}{2!}+ \frac{1+3+3^2}{3!}+....... $
The series becomes $ \sum_{k=1}^{\infty} \frac {3^{k-1}}{k!}$.
How to calculate it's sum? Is it divergent due to the numerator?
 [/INST] The numerator of the $k$th term: $1+3+9+\cdots+3^{k-1}=\dfrac{3^k-1}{2}$
The denominator of the $k$th term: $k!$
The sum:
$$\sum_{k=1}^\infty \dfrac{3^k-1}{2\left(k!\right)}=\dfrac{1}{2}\sum_{k=1}^\infty \dfrac{3^k}{k!}-\dfrac{1}{2}\sum_{k=1}^\infty \dfrac{1}{k!}=\dfrac{1}{2}\left(e^3-1\right)-\dfrac{1}{2}\left(e-1\right)=\dfrac{e^3-e}{2}$$
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