<s>[INST] Sum of infinite series by considering Maclaurin Series for $e^x$ I want to find:
$ \sum_{n = 0}^{ \infty } \frac{1505n + 1506}{3^n(n+1)!}$
We have:
$e^\frac{x}{3} = \sum_{n = 0}^{ \infty } \frac{x^n}{3^n(n!)}$
which we could integrate on both sides to get:
$ \int e^\frac{x}{3} dx =  \sum_{n = 0}^{ \infty } \frac{ x^{n+1} }{3^n({n+1}!)}$
Now, I'm just missing the numerator $1505n+1506$, which I cannot for the life of me see how to proceed... Any hints would be appreciated!
$\textbf{Problem resolved using hint provided below:}$
$\sum_{n = 0}^{ \infty } \frac{1505n + 1506}{3^n(n+1)!} = \sum_{n = 0}^{ \infty } \frac{1505(n+1) + 1}{3^n(n+1)!} = \sum_{n = 0}^{ \infty } \frac{1505}{3^nn!} + \sum_{n = 0}^{ \infty } \frac{1}{3^n(n+1)!}$
The first term can be computed using:
$e^\frac{x}{3} = \sum_{n = 0}^{ \infty } \frac{x^n}{3^nn!}$
, where we set $x = 1$ to get: $1505e^\frac{1}{3}$
The second term can be computed by taking the definte integral of $e^\frac{x}{3}$:
$ \intop\nolimits_{0}^{1} e^\frac{x}{3} dx =  \sum_{n = 0}^{ \infty } \frac{ 1^{n+1} }{3^n({n+1}!)} - \frac{ 0^{n+1} }{3^n({n+1}!)} = \sum_{n = 0}^{ \infty } \frac{1}{3^n{(n+1)}!} = 3e^\frac{1}{3}-3$
Sum = $1508e^\frac{1}{3}-3$
 [/INST] $$S=\sum_{n=0}^{\infty} \frac{an+b}{3^n (n+1)!}= \sum_{n=0}^{\infty} \frac{(a(n+1)+b-a}{3^n (n+1)!}=a\sum_{n=0}^{n}\frac{3^{-n }}{n!}+(b-a)\sum_{n=0}^{\infty} \frac{3^{-n}}{(n+1)!}$$
$$\implies S=ae^{-3}+3(b-a)\sum_{n=0}^{\infty}\frac{3^{-(n+1)}}{(n+1)!}=ae^{-3}+3(b-a)[e^{-3}-1].$$
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