Finding $\sum_{k=1}^{\infty}k^2 \frac{2^{k-1}}{3^k}$ Finding $$\sum_{k=1}^{\infty}k^2 \frac{2^{k-1}}{3^k}$$
I think if $$\int_{1}^{\infty}x^2 \frac{2^{x-1}}{3^x}dx$$
 exists that this sequence is convergent, but I doubt that this integral is equal to the number to which the sequence converges to. I do not know a way to solve this sequence, thought about doing geometric progression, but can't because of $k^2$ 
 A: We know that the geometric series converges,
$$\sum_{k = 0}^{\infty} x^k = \frac{1}{1 - x} \;\; |x| < 1$$
Differentiating both sides,
$$\sum_{k = 0}^{\infty} kx^{k-1} = \frac{1}{(1 - x)^2} \;\; |x| < 1$$
Therefore, multiplying throughout by $x$ we get,
$$\sum_{k = 0}^{\infty} kx^{k} = \frac{x}{(1 - x)^2} \;\; |x| < 1$$
Differentiating again we get,
$$\sum_{k = 0}^{\infty} k^2x^{k-1} = \frac{1 + x}{(1 - x)^3} \;\; |x| < 1$$
Multiplying throughout by $x$ we get,
$$\sum_{k = 0}^{\infty} k^2x^{k} = \frac{x(1 + x)}{(1 - x)^3} \;\; |x| < 1$$
Substitute $x = \frac{2}{3} < 1$ to get,
$$\sum_{k = 0}^{\infty} k^2\left(\frac{2}{3}\right)^{k} = 30$$
Divide throughout by $2$ to get,
$$\sum_{k = 0}^{\infty} k^2\frac{2^{k-1}}{3^k} = 15$$
A: Let $f(x)$ be given by 
$$f(x)=\sum_{k=0}^\infty x^k=\frac1{1-x}$$
for $-1\le x<1$.  If we differentiate $f$ we find that 
$$f'(x)=\sum_{k=0}^\infty kx^{k-1}=\frac1{(1-x)^2} \tag 1$$
Multiplying $(1)$ by $x$ and differentiating again yields
$$(xf'(x))'=\sum_{k=0}^\infty k^2x^{k-1}=\frac{1+x}{(1-x)^3} \tag 2$$
Now, letting $x=2/3$ in $(2)$ reveals
$$\sum_{k=1}^\infty k^2\frac{2^{k-1}}{3^{k-1}}=3\sum_{k=1}^\infty k^2\frac{2^{k-1}}{3^{k-1}}=\frac{5/3}{(1/3)^3}$$
Finally, we can write
$$\sum_{k=1}^\infty k^2\frac{2^{k-1}}{3^{k-1}}=15$$
A: $\sum_{k=1}^{\infty}k^2 \frac{2^{k-1}}{3^k}$
Start with $\sum_{k=0}^{\infty} x^k=\frac1{1-x}$.
Take the derivative $\sum_{k=1}^{\infty} kx^{k-1}=\frac{1}{(1-x)^2}$.
Now multiply both sides by $x$ to get 
$\sum_{k=1}^{\infty} kx^{k}=\frac{x}{(1-x)^2}$.
Now take the derivative again and multiply by $x$ again to get
$\sum_{k=1}^{\infty} k^2x^{k}=\frac{x(1+x)}{(1-x)^3}$.
Now divide by $2$ and plug in $x=2/3$ and you'll have the answer.
I believe it  equals $15$.
A: This is the same as
$$ \frac{1}{2} \sum_{k \geq 1} k^2 \left( \frac{2}{3} \right)^k,$$
which looks almost like the geometric series
$$ f(x) = \frac{1}{2} \sum_{k \geq 1} x^k = \frac{1}{2} \frac{x}{1 - x}$$
at $x = \frac{2}{3}$. Notice that
$$ \frac{d}{dx} f(x) = \frac{1}{2} \sum_{k \geq 1} k x^{k - 1},$$
which is closer to what we want. We want another factor of $k$, so we multiply by $x$ and differentiate again,
$$ \frac{d}{dx} x \frac{d}{dx} f(x) = \frac{1}{2} \sum_{k \geq 1} k^2 x^{k-1}.$$
We are only missing a single factor of $x$, so we multiply by $x$ again to see that
$$ \left( x \frac{d}{dx} \right)^2 f(x) = \frac{1}{2} \sum_{k \geq 1} k^2 x^k.$$
To evaluate, you need only to compute
$$ \left( x \frac{d}{dx} \right)^2 \frac{1}{2} \frac{x}{1 - x}$$
and evaluate it at $x = \frac{2}{3}$.
A: The sequence can be written as $=\frac{1}{2}\sum_{1}^{\infty} k^2.a^k$ where $a=\frac{2}{3}$
Let $$S = \sum_{1}^{\infty} k^2.a^k\tag{1}$$
Multiply (1) by a,
$$aS = a^2+4a^3+9a^4+\cdot+\infty$$
Subtract now 
$$(1-a)S = a+3a^2+5a^3+\cdot+\infty = 2a-a + 4a^2-a^2 + 6a^3-a^3 +\cdot+\infty$$
$$(1-a)S = 2a(1+2a+3a^2+\cdot+\infty) - a(1+a+a^2+\cdot+\infty)$$
$$(1-a)S = 2aS'-\frac{a}{1-a}$$
$$S' = 1+2a+3a^2+4a^3+\cdot+\infty$$
$$aS' = a+2a^2+3a^3+\cdot+\infty$$
$$(1-a)S' = 1+a+a^2+a^3+\cdot+\infty$$
$$S' = \frac{1}{(1-a)^2}$$
$$(1-a)S = \frac{2a}{(1-a)^2} - \frac{a}{1-a}$$
$$S = \frac{2a}{(1-a)^3} - \frac{a}{(1-a)^2}$$
Substituting the value of $a = \frac{2}{3}$
The sequence equals $\frac{1}{2}\sum_{1}^{\infty} k^2.a^k= \boxed{15}$
