eliaskidszs

2022-01-06

Using power series representation, calculate

$\sum _{n=1}^{\mathrm{\infty}}\frac{n{2}^{n}}{{3}^{n}}$

Jim Hunt

Beginner2022-01-07Added 45 answers

Recall that, in general,

$1+x+{x}^{2}+\dots =\frac{1}{1-x},\text{}\left|x\right|1$

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

$1+2x+3{x}^{2}+\dots =\frac{1}{{(1-x)}^{2}},\text{}\left|x\right|1$

Now, multiplying both sides by x leads to

$x+2{x}^{2}+3{x}^{3}+\dots =\sum _{n=1}^{\mathrm{\infty}}n{x}^{n}=\frac{x}{{(1-x)}^{2}},\text{}\left|x\right|1$

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

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

Marcus Herman

Beginner2022-01-08Added 41 answers

Basically you start with

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

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

$\sum _{x=1}^{\mathrm{\infty}}n{x}^{n-1}=\frac{1}{{(1-x)}^{2}}$

A popular second step you can do from there is multiply both sides by x, which gives you

$\sum _{x=1}^{\mathrm{\infty}}n{x}^{n}=\frac{x}{{(1-x)}^{2}}$

And then you do all the mathematical operations such as

A popular second step you can do from there is multiply both sides by x, which gives you

karton

Expert2022-01-11Added 613 answers

First observe that your series is the special case of

with

By using the Cauchy product on

and after multiplying by z

For

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