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

eliaskidszs 2022-01-06 Answered
Using power series representation, calculate
\(\displaystyle{\sum_{{{n}={1}}}^{\infty}}{\frac{{{n}{2}^{{n}}}}{{{3}^{{n}}}}}\)

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Expert Answer

Jim Hunt
Answered 2022-01-07 Author has 2003 answers
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.
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Marcus Herman
Answered 2022-01-08 Author has 1947 answers
Basically you start with
\(\displaystyle{\sum_{{{x}={0}}}^{\infty}}{x}^{{n}}={\frac{{{1}}}{{{1}-{x}}}}\)
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
\(\displaystyle{\sum_{{{x}={1}}}^{\infty}}{n}{x}^{{n}}={\frac{{{x}}}{{{\left({1}-{x}\right)}^{{2}}}}}\)
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karton
Answered 2022-01-11 Author has 8659 answers

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 \(\frac{\frac{2}{3}}{(1-\frac{2}{3})^2}=2\cdot3=6\)

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