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I want to find an explicit formula for $\sum_{n=0}^\infty n^3x^n$ for $|x|\le1$.Is the idea that first to show that this series is convergent and then we can find the number that it converges to? I tried to use ratio test, but it didn't work. Any suggestion? Thanks!

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  • $\begingroup$ If $x=1$, this series diverges. $\endgroup$ – Andrew Apr 24 '16 at 19:59
  • $\begingroup$ When $|x|<1$, ratio test works!!! $\endgroup$ – Mambo Apr 24 '16 at 19:59
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    $\begingroup$ Hint $$n^3 =(n+1)n(n-1) +n $$ $\endgroup$ – MotylaNogaTomkaMazura Apr 24 '16 at 20:00
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The Ratio Test will tell you for what $x$ the series converges:

$$L = \lim_{n\to\infty} \left|\frac{(n+1)^3x^{n+1}}{n^3x^n}\right| = |x|\lim_{n\to\infty} \frac{(n+1)^3}{n^3} = |x|$$

And the ratio test tells us that the series converges absolutely if $|x|<1$ (you should check that it diverges at the endpoints).

To compute its value, first notice $$n^3 = n(n-1)(n-2) + 3n(n-1) + n$$

Next, we start with the geometric series: $$S_0=\sum_{n=0}^\infty x^n = \frac{1}{1-x},\quad |x|<1$$ Differentiate termwise to get: $$\sum_{n=1}^\infty n x^{n-1} = \frac{1}{(1-x)^2},\quad |x|<1$$ Multiply by $x$ to get $$S_1 = \sum_{n=1}^\infty nx^n = \frac{x}{(1-x)^2},\quad |x|<1$$ Repeat two more times to get: $$S_2=\sum_{n=2}^\infty n(n-1) x^{n} = \frac{2x^2}{(1-x)^3},\quad |x|<1$$

$$S_3=\sum_{n=3}^\infty n(n-1)(n-2) x^{n} = \frac{6x^3}{(1-x)^4},\quad |x|<1$$

See if you can use the above information to find a combination of these that gives $\sum_{n=0}^\infty n^3x^n$.


By the way, even though the bottom indeces on $S_1$, $S_2$, $S_3$ are not $0$, we can simply change them all to $0$, because we're adding multiple copies of $0$. (We change indeces at every step, since the constant terms vanish rather than introducing negative powers of $x$. But in this case, there's nothing to worry about.)

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  • $\begingroup$ Since $n^3=(n+1)n(n-1)+n$, i can just use $S_2$ and multiply it with $(n+1)$ and then plus $nx^n$? $\endgroup$ – J.doe Apr 24 '16 at 23:07
  • $\begingroup$ The trouble with that is you are summing over $n$, so it's unclear how you define $(n+1)S_2$ (since $S_2$ does not depend on $n$, the question would be what $n$ you multiply $S_2$ by in order to get the desired result). You would be better suited trying (for instance) $$S_3 + 3S_2 + S_1$$ at which point your only task becomes convincing yourself that this does indeed give $\sum_{n=0}^\infty n^3 x^n$. $\endgroup$ – Nicholas Stull Apr 25 '16 at 1:00
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    $\begingroup$ When I saw $\sum_{n=0}^\infty n^3x^n$, I immediately thought of derivatives of $\sum_{n=0}^\infty x^n$. The first derivative brings $n$ down, the next $n-1$, the next $n-2$, and so on. Keep going until you have $n^3$ as the leading order of $n$ in $S_{k}$ (in this case, $S_3$). Then, you recognize that $n(n-1)(n-2)$ is taking away $n^2$ terms, which you must compensate for with $n(n-1)$ terms, and you repeat. $\endgroup$ – Nicholas Stull Apr 25 '16 at 1:50
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    $\begingroup$ The only other thing I needed is that when I take derivatives, I lose powers of $x$. So while I take 3 derivatives of $\sum_{n=0}^\infty x^n$, I also recognize that to get back to $x^n$, I need to multiply by the 3 powers of $x$ I have lost in the process of taking derivatives. $\endgroup$ – Nicholas Stull Apr 25 '16 at 1:51
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    $\begingroup$ You're welcome. I'm just glad I was able to help. $\endgroup$ – Nicholas Stull Apr 25 '16 at 2:28
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By the root test $$\limsup_{n\to\infty} \sqrt[n]{|n^3x^n|}=|x|\limsup_{n\to\infty}n^{3/n}=|x|$$ so we have convergence for $|x|<1$.

Now note that $$\sum_{n=0}^\infty x^n=\frac{1}{1-x}$$ so $$\sum_{n=0}^\infty nx^{n-1}=\frac{1}{(1-x)^2}$$ and $$\sum_{n=0}^\infty nx^{n}=\frac{x}{(1-x)^2}.$$ You should be able to continue this process.

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  • $\begingroup$ You have a sign error in the last two series. $\endgroup$ – Nicholas Stull Apr 24 '16 at 20:20
  • $\begingroup$ They always get me. Fixed. $\endgroup$ – Elliot G Apr 24 '16 at 20:21
  • $\begingroup$ It took me a bit of staring to catch it myself (since they get me all the time too). +1 of course. $\endgroup$ – Nicholas Stull Apr 24 '16 at 20:38
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For $x = \pm 1$, you can find the value of the series separately and its not that hard. For $|x| < 1$, consider $f(x) = \displaystyle \sum_{n=0}^\infty x^n= \dfrac{1}{1-x}$, then find $xf'(x) = \displaystyle \sum_{n=0}^\infty nx^n= x(1-x)^{-2}$,and repeat this until you get to the desire series.

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