# Using power series representation, calculate \sum_{n=1}^\infty \frac{n2^n}{3^n}

Using power series representation, calculate
$$\displaystyle{\sum_{{{n}={1}}}^{\infty}}{\frac{{{n}{2}^{{n}}}}{{{3}^{{n}}}}}$$

• Questions are typically answered in as fast as 30 minutes

### Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Jim Hunt
Recall that, in general,
$$\displaystyle{1}+{x}+{x}^{{2}}+\ldots={\frac{{{1}}}{{{1}-{x}}}},\ {\left|{x}\right|}{ < }{1}$$
Moreover, power series can be differentiated term by term. So, differentiating both sides of the equation above we get
$$\displaystyle{1}+{2}{x}+{3}{x}^{{2}}+\ldots={\frac{{{1}}}{{{\left({1}-{x}\right)}^{{2}}}}},\ {\left|{x}\right|}{ < }{1}$$
Now, multiplying both sides by x leads to
$$\displaystyle{x}+{2}{x}^{{2}}+{3}{x}^{{3}}+\ldots={\sum_{{{n}={1}}}^{\infty}}{n}{x}^{{n}}={\frac{{{x}}}{{{\left({1}-{x}\right)}^{{2}}}}},\ {\left|{x}\right|}{ < }{1}$$
However, in this case $$\displaystyle{x}=\frac{{2}}{{3}}{ < }{1}$$, so simply substitute $$\displaystyle{x}=\frac{{2}}{{3}}$$ in formula above.
###### Not exactly what you’re looking for?
Marcus Herman
$$\displaystyle{\sum_{{{x}={0}}}^{\infty}}{x}^{{n}}={\frac{{{1}}}{{{1}-{x}}}}$$
And then you do all the mathematical operations such as $$\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}$$ on both sides until you get the form you want. For example, the first derivative will give you
$$\displaystyle{\sum_{{{x}={1}}}^{\infty}}{n}{x}^{{{n}-{1}}}={\frac{{{1}}}{{{\left({1}-{x}\right)}^{{2}}}}}$$
A popular second step you can do from there is multiply both sides by x, which gives you
$$\displaystyle{\sum_{{{x}={1}}}^{\infty}}{n}{x}^{{n}}={\frac{{{x}}}{{{\left({1}-{x}\right)}^{{2}}}}}$$
karton

First observe that your series is the special case of
$$\sum_{n=1}^\infty nz^n$$
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
$$(\frac{1}{1-z})^2=(\sum_{n=0}^\infty z^n)^2=\sum_{n=0}^\infty(\sum_{k=0}^n)z^n=\sum_{n=0}^\infty(n+1)z^n=\sum_{n=1}^\infty nz^{n-1}$$
and after multiplying by z
$$\sum_{n=1}^\infty nz^n=\frac{z}{(1-z)^2}$$
For $$z=\frac{2}{3}$$ we get $$\frac{\frac{2}{3}}{(1-\frac{2}{3})^2}=2\cdot3=6$$