I am looking for a way to derive the partial sum formula for $ \large \sum_{n=1}^{k}\frac{n^3}{3^n}$. I notice that the wolframalpha website evaluates it as $ \frac 1 8 \frac{-4 k^3-18 k^2-36 k+33 (3^k-1) }{3^k} $ but there is no indication how it derived this.


In general I am looking for a way to find a closed form expression for $ f(x) = \large \sum_{n=1}^{k}x^n n^m$ where $m$ is a positive integer and $x$ is a real number. I am looking at easier cases first to see if a general pattern emerges.

Using the suggestion in the comments, using the geometric sum formula we can differentiate both sides of the expression

$ \large \sum_{n=0}^{k-1} x ^n = \frac{1-x^k}{1-x}$ whence we derive $ \large \sum_{n=0}^{k-1} nx ^{n-1} = \frac{(k-1)x^k - kx^{k-1}+1}{(1-x)^2}$

Then multiplying both sides by $x$ will give us a closed form expression for $ \sum_{n=0}^{k-1} nx ^{n}$. Now we can take the derivative again.

When you take the derivative of a sum does the upper limit and lower limit change? I did read once that you can increase the lower limit by 1 each time. It looks like the $n=0$ term vanishes here anyway so there is no harm in starting index at $0$.

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    $\begingroup$ Start with the formula for the geometric sum, then differentiate both sides twice. $\endgroup$ – Ahmed S. Attaalla Oct 15 '16 at 19:16
  • $\begingroup$ Thanks for replying. I updated the question to make it harder. Should I differentiate it three times? $\endgroup$ – john Oct 15 '16 at 19:28
  • $\begingroup$ @AhmedS.Attaalla This method gives the infinite sum, but how do we get the partial sums ? $\endgroup$ – Peter Oct 15 '16 at 19:34
  • $\begingroup$ @Peter the formula for the geometric sum $\sum_{n < k} z^n = \frac{z^k-1}{z-1}$ $\endgroup$ – reuns Oct 15 '16 at 20:02
  • $\begingroup$ And we derive this, OK, thank you. $\endgroup$ – Peter Oct 15 '16 at 20:04

As an alternative to Ahmed's proposal, you may compute $$ T_k(x) = \sum_{n=1}^{k}\frac{e^{nx}}{3^n} $$ then evaluate $T_k'''(0)$, since $$ T_k'''(x) = \sum_{n=1}^{k}\frac{n^3 e^{nx}}{3^n}. $$

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$$\sum_{i =0}^{k-1} x^{i }=\frac{1-x^{k}}{1-x}$$

$$\sum_{i =0}^{k-1} \frac{\partial}{\partial x} x^i=\frac{\partial}{\partial x} \frac{1-x^{k}}{1-x}$$

Using the technique you mentioned we have:

$$\sum_{i =0}^{k-1} ix^{i}=x\frac{\partial}{\partial x} \frac{1-x^{k}}{1-x}$$

Now a second time,

$$\sum_{i=0}^{k-1} i^2x^{i}=x\frac{\partial}{\partial x}\left(x\frac{\partial}{\partial x} \frac{1-x^{k}}{1-x}\right)$$

A third time,

$$\sum_{i=0}^{k-1} i^3x^i=x\frac{\partial}{\partial x}\left(x\frac{\partial}{\partial x}\left(x\frac{\partial}{\partial x} \frac{1-x^{k}}{1-x}\right)\right)$$

And so on.

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