I hope this hasn't been asked already, though I have looked around the site and found many similar answers.


Form Cauchy product of two series: $a_k\;x^k$ and $\tfrac{1}{1-x}=1+x+x^2+\cdots$.

So I come up with, $\sum_{n=0}^{\infty}\;c_n = \sum_{n=0}^{\infty}\;\sum_{k+l=n}\;a_l\;b_k = \cdots = \sum_{n=0}^{\infty}\;x^n\;\sum_{k=0}^n\;a_k = x^0\;(a_0)+x^1\;(a_0+a_1)+x^2\;(a_0+a_1+a_2)+\cdots$.

It asks for what values of $x$ this would be valid:

This is a funny question to me because it depends upon the coefficients in the power series, right? If I take the ratio test, I get $\lim_{n\to\infty}\;\bigg| \frac{a_{k+1}\;x^{k+1}}{a_k\;x^k}\bigg| = |x|\cdot \lim_{k\to\infty}\;\big| \frac{a_{k+1}}{a_k} \big|$. For this series to be convergent, doesn't this have to come to a real number, $L$ (not in $\mathbb{\bar{R}}$)? Therefore, $|x|<1/L$?

I know that the other series, $\sum_{n=0}^{\infty}\;x^n$ converges for $r \in (-1,1)$.

So for the product to be convergent, doesn't the requirement of $x$ have to be, $|x| < \min\{1,1/L\}$?

The reason I include this, other than the questions above, is that the question suggests using "this approach" to attain a closed form $\sum_{k=0}^{\infty}\;k\;x^k$, for $x \in (-1,1)$. By using the ratio test (my favorite), I'm pretty sure that for this to converge, $|x|<1$ - which is given. I tried writing out some of the terms but they do not seem to reach a point whereby future terms cancel (as they do in a series like $\sum_{n=1}^{\infty} \frac{1}{k\;(k+2)\;(k+4)}$). I've tried bounding (Squeeze) them but didn't get very far.

Thanks for any suggestions!

  • $\begingroup$ Well I can't think of any other "approach" to use other than reverting back to things like integrating, differentiating and the usual. Thanks anyways. $\endgroup$ – nate Mar 9 '12 at 8:12

Note that $\sum_{n\ge 0}nx^n$ is almost the Cauchy product of $\sum_{n\ge 0}x^n$ with itself: that Cauchy product is

$$\left(\sum_{n\ge 0}x^n\right)^2=\sum_{n\ge 0}x^n\sum_{k=0}^n 1^2=\sum_{n\ge 0}(n+1)x^n\;.\tag{1}$$

If you multiply the Cauchy product in $(1)$ by $x$, you get

$$x\sum_{n\ge 0}(n+1)x^n=\sum_{n\ge 0}(n+1)x^{n+1}=\sum_{n\ge 1}nx^n=\sum_{n\ge 0}nx^n\;,\tag{2}$$

since the $n=0$ term is $0$ anyway. Combining $(1)$ and $(2)$, we have

$$\sum_{n\ge 0}nx^n=x\left(\sum_{n\ge 0}x^n\right)^2=x\left(\frac1{1-x}\right)^2=\frac{x}{(1-x)^2}\;,$$

with convergence for $|x|<1$ by the reasoning that you gave.

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  • $\begingroup$ Hi @Brian, thanks for the help! Now I understand what the "approach" was - having that "nested" summation on the inside. I hadn't seen that before. I followed your logic well except for the last line. How did you get from the first term to the second? $\endgroup$ – nate Mar 9 '12 at 15:18
  • $\begingroup$ @nate: The first term is $x$ times the Cauchy product of $\sum_{n\ge 0}x^n$ with itself, i.e., $x\left(\sum_nx^n\right)^2$. I’ve revised the answer a little to make this clearer; take a look. $\endgroup$ – Brian M. Scott Mar 9 '12 at 15:33
  • $\begingroup$ Okay, went through it again and it now makes sense to me. Is the first term at the very top supposed to be $\sum_{n\geq 0} n\;x^n$? And was it arbitrary to choose $1^2$ in $(1)$? Could it just as easily been $1^7$, or $1^1$? $\endgroup$ – nate Mar 9 '12 at 16:04
  • $\begingroup$ @nate: The $k$ was a typo; I’ve fixed it now. $1^2$ was not arbitrary: the coefficient of $x^n$ in the Cauchy product of $\sum_n a_n$ and $\sum_n b_n$ is $\sum_{k=0}^n a_kb_k$; in this particular case $a_k=b_k=1$, so $a_kb_k=1^2$. $\endgroup$ – Brian M. Scott Mar 9 '12 at 16:08
  • $\begingroup$ Oh, I can see that. Okay, well I really appreciate your time and help! Thanks again. $\endgroup$ – nate Mar 9 '12 at 16:10

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