Using power series representation, calculate ∑n=1∞n2n3n

Question

Using power series representation, calculate  ∑n=1∞n2n3n

Jim Hunt · Accepted Answer

Recall that, in general,  1+x+x2+…=11−x, |x|&amp;lt;1  Moreover, power series can be differentiated term by term. So, differentiating both sides of the equation above we get  1+2x+3x2+…=1(1−x)2, |x|&amp;lt;1  Now, multiplying both sides by x leads to  x+2x2+3x3+…=∑n=1∞nxn=x(1−x)2, |x|&amp;lt;1  However, in this case x=23&amp;lt;1, so simply substitute x=23 in formula above.

Marcus Herman · Accepted Answer

Basically you start with  ∑x=0∞xn=11−x  And then you do all the mathematical operations such as ddx on both sides until you get the form you want. For example, the first derivative will give you  ∑x=1∞nxn−1=1(1−x)2  A popular second step you can do from there is multiply both sides by x, which gives you  ∑x=1∞nxn=x(1−x)2

karton · Accepted Answer

First observe that your series is the special case of ∑n=1∞nzn with z=23, which has radius of convergence R=1 By using the Cauchy product on ∑n=0∞zn=11−z we get (11−z)2=(∑n=0∞zn)2=∑n=0∞(∑k=0n)zn=∑n=0∞(n+1)zn=∑n=1∞nzn−1 and after multiplying by z ∑n=1∞nzn=z(1−z)2 For z=23 we get 23(1−23)2=2⋅3=6

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

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