2022-01-06

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

Jim Hunt

Expert

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

Expert

$\sum _{x=0}^{\mathrm{\infty }}{x}^{n}=\frac{1}{1-x}$
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

Expert

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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