Power series representation/calculation I am struggling a bit with power series at the moment, and I don't quite understand what this question is asking me to do? Am I meant to form a power series from these, or simply evaluate that series? Any explanation/working is appreciated.
Using power series representation, calculate
$$\sum_{n=1}^\infty \frac{n2^n}{3^n}.$$ 
 A: Recall that, in general,
$$1 + x + x^2 + \cdots = \frac{1}{1 - x}, \quad |x| < 1.$$
Moreover, power series can be differentiated term by term. So, differentiating both sides of the equation above we get
$$1 + 2x + 3x^2 + \cdots = \frac{1}{(1 - x)^2}, \quad |x| < 1.$$
Now, multiplying both sides by $x$ leads to
$$x + 2x^2 + 3x^3 + \cdots = \sum_{n = 1}^\infty nx^n = \frac{x}{(1 - x)^2}, \quad |x| < 1.$$
However, in this case $x = 2/3 < 1$, so simply substitute $x = 2/3$ in formula above.
A: Hint: Define $f$ by
$$f(x) = \sum_{n = 1}^{\infty} x^n$$
Using the formula for a geometric series, this is the same as
$$f(x) = \frac{x}{1 - x}$$
Now compute $f'$ two different ways.
A: I believe this Youtube video does a great job explaining the issue. Basically you start with
$$ \sum_{x = 0}^{\infty} x ^ n = \frac{1}{1 - x}.$$
And then you do all the mathematical operations such as $ \frac{d}{dx} $ on both sides until you get the form you want. For example, the first derivative will give you
$$ \sum_{x = 1}^{\infty} n x ^ {n-1} = \frac{1}{(1 - x) ^ 2}.$$
A popular second step you can do from there is multiply both sides by $x$, which gives you
$$ \sum_{x = 1}^{\infty} n x ^ {n} = \frac{x}{(1 - x) ^ 2}.$$
A: First observe that your series is the special case of
$$\sum_{n=1}^\infty n z^n$$
with $z = \frac{2}{3}$, which has radius of convergence $R=1$.
By using the Cauchy product on $\sum_{n=0}^\infty z^n = \frac{1}{1-z}$ we get
$$\big( \frac{1}{1-z} \big)^2 = \Big( \sum_{n=0}^\infty z^n \Big)^2 = \sum_{n=0}^\infty \Big( \sum_{k=0}^n \Big) z^n = \sum_{n=0}^\infty (n+1) z^n = \sum_{n=1}^\infty n z^{n-1}$$
and after multiplying by $z$
$$\sum_{n=1}^\infty n z^n = \frac{z}{(1-z)^2}$$
For $z = \frac{2}{3}$ we get $\frac{\frac{2}{3}}{(1-\frac{2}{3})^2} = 2 \cdot 3 = 6$.
