I have this math problem and I'm not entirely sure how to set it up.
It is know that $\sum_{n=1}^{\infty}=\frac{n}{2^n}=2$ and $\sum_{n=1}^{\infty}=\frac{1}{n2^n}=ln(2)$. Supposing that $a, b, $and $c$ are constants, evaluate $\sum_{n=1}^{\infty}\frac{an^2+cn+b}{n2^n}$.
I know I have to use both the sums that were given, however I'm not sure how to set it up.
HINT: Because the sums converge (separately) you can sum them and get the same result
HINT $2$: If $\underset{{n \to \infty}}{\lim} b_n = b$ and $\lim_{n \to \infty} a_n = a$ then $$\lim_{n \to \infty} b_n + a_n = \lim_{n \to \infty} a_n + \lim_{n \to \infty} b_n = a + b$$
Sol: $$\sum_{n=1}^{\infty}\frac{an^2+cn+b}{n2^n} = \sum_{n=1}^{\infty} \dfrac{an^2}{n2^n} + \sum_{n=1}^{\infty} \dfrac{cn}{n2^n} + \sum_{n=1}^{\infty} \dfrac{b}{n2^n} = 2a + c -1 + b\log(2)$$