2022-01-06

Using power series representation, calculate
$\sum _{n=1}^{\mathrm{\infty }}\frac{n{2}^{n}}{{3}^{n}}$

Jim Hunt

Recall that, in general,

Moreover, power series can be differentiated term by term. So, differentiating both sides of the equation above we get

Now, multiplying both sides by x leads to

However, in this case $x=\frac{2}{3}<1$, so simply substitute $x=\frac{2}{3}$ in formula above.

Marcus Herman

$\sum _{x=0}^{\mathrm{\infty }}{x}^{n}=\frac{1}{1-x}$
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
$\sum _{x=1}^{\mathrm{\infty }}n{x}^{n-1}=\frac{1}{{\left(1-x\right)}^{2}}$
A popular second step you can do from there is multiply both sides by x, which gives you
$\sum _{x=1}^{\mathrm{\infty }}n{x}^{n}=\frac{x}{{\left(1-x\right)}^{2}}$

karton

First observe that your series is the special case of
$\sum _{n=1}^{\mathrm{\infty }}n{z}^{n}$
with $z=\frac{2}{3}$, which has radius of convergence R=1
By using the Cauchy product on $\sum _{n=0}^{\mathrm{\infty }}{z}^{n}=\frac{1}{1-z}$ we get
$\left(\frac{1}{1-z}{\right)}^{2}=\left(\sum _{n=0}^{\mathrm{\infty }}{z}^{n}{\right)}^{2}=\sum _{n=0}^{\mathrm{\infty }}\left(\sum _{k=0}^{n}\right){z}^{n}=\sum _{n=0}^{\mathrm{\infty }}\left(n+1\right){z}^{n}=\sum _{n=1}^{\mathrm{\infty }}n{z}^{n-1}$
and after multiplying by z
$\sum _{n=1}^{\mathrm{\infty }}n{z}^{n}=\frac{z}{\left(1-z{\right)}^{2}}$
For $z=\frac{2}{3}$ we get $\frac{\frac{2}{3}}{\left(1-\frac{2}{3}{\right)}^{2}}=2\cdot 3=6$

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