How to compute $\sum_{n=0}^{\infty}{\frac{n(n+2)}{3^n}}$? so I am having problems finding the value of $\sum_{n=0}^{\infty}{\frac{n(n+2)}{3^n}}$. WolframAlpha says it should be $=3$, but what's the reasoning behind it? Can I just look at $1/3^n$ since the series is absolutetly convergent and we can disregard some terms...
 A: The following may be useful.
Let $F(x)=1+x+x^2+x^3+\cdots$, which equals $\displaystyle {1\over1-x}$ if $|x|<1$. Then also for $|x|<1$, $$F'(x)=1+2x+3x^2+4x^3+\cdots$$
$${F'(x)-1\over x}=2+3x+4x^2+5x^3\cdots$$
$${d\over dx}{F'(x)-1\over x}=(1\cdot3)+(2\cdot4)x+(3\cdot5)x^2+\cdots$$
$$x{d\over dx}{F'(x)-1\over x}=(1\cdot3)x+(2\cdot4)x^2+(3\cdot5)x^3+\cdots$$
If you evaluate both sides of the final equation at $x={1\over3}$, you will find your answer.
A: Telescopic approach. Note that
$$\frac{n(n+2)}{3^n}=a_n-a_{n+1}$$
where
$$a_n=\frac{(n+1)(n+2)}{2\cdot 3^{n-1}}.$$
Hence
$$\sum_{n=0}^{N-1}{\frac{n(n+2)}{3^n}}=(a_0-a_1)+(a_1-a_2)+\dots+(a_{N-1}-a_N)=a_0-a_{N}=3-a_N$$
and it follows that
$$\sum_{n=0}^{\infty}{\frac{n(n+2)}{3^n}}=3-\lim_{N\to \infty}a_N=3-0=3.$$
A: Let $$\dfrac{n(n+2)}{3^n}=f(n+1)-f(n)$$ to form a https://en.m.wikipedia.org/wiki/Telescoping_series
where $f(m)3^m=a_0+a_1m+a_2m^2+a_3m^3+\cdots$
$3n(n+2)=a_0+a_1(n+1)+a_2(n+1)^2+\cdots$
$-3(a_0+a_1n+a_2n^2+\cdots)$
Comparing the coefficients of $n^2,$ $$3=a_2-3a_2\iff a_2=?$$
Compare the coefficients of $n,$ $$6=2a_2+a_1-3a_1\implies a_1=?$$
Compare the constant, $$0=a_0+a_1+a_2-3(a_0+a_1+a_2)\implies a_0=?$$
Of course, we need to establish $$\lim_{n\to\infty}f(n)=0$$
