1
$\begingroup$

The form of the summation I have is $$\sum _{ x=0 }^{ \infty }{ x{ a }^{ x } } $$ I need to somehow remove the $x$ from the original summation in order to achieve the geometric series in each other summation. For instance, $$\sum _{ x=? }^{ \infty }{ { a }^{ x }\quad +\quad } \sum _{ x=? }^{ \infty }{ { a }^{ x }\quad +\quad } \sum _{ x=? }^{ \infty }{ { a }^{ x }\quad +\quad } ...$$ I have seen this done before, but forget how to work with the bounds of each new summation . It would be greatly appreciated if anyone had a clue what I was talking about.

$\endgroup$
3
$\begingroup$

You should know that $\sum_{n=0}^\infty a^n = \frac{1}{1-a}$.

If we were to derive both sides with respect to $a$, then we have

$\frac{d}{da}[\sum_{n=0}^\infty a^n] = \sum_{n=0}^\infty \frac{d}{da}[a^n]=\sum_{n=0}^\infty na^{n-1} = \frac{d}{da}[\frac{1}{1-a}]=\frac{1}{(1-a)^2}$.

Multiplying both sides by $a$ gives us the desired result:

$$\sum_{n=0}^\infty na^n = \frac{a}{(1-a)^2}$$

$\endgroup$
2
$\begingroup$

Here is the way forward that aligns with approach suggested in the OP.

$$\begin{align} \sum_{x=0}^\infty xa^x&=\sum_{x=1}^\infty a^x\sum_{y=1}^x(1)\\\\ &=\sum_{y=1}^\infty\sum_{x=y}^{\infty}a^x\\\\ & =\sum_{y=1}^\infty\frac{a^y}{1-a}\\\\ &=\frac{a}{(1-a)^2} \end{align}$$

$\endgroup$
  • $\begingroup$ I believe that I understand your question. Please let me know how I can improve my answer. I really just want to give you the best answer I can. - Mark $\endgroup$ – Mark Viola Feb 25 '16 at 4:42
0
$\begingroup$

$a^x=e^{x\log a}$ and then apply the Taylor Expansion for $e^{x\log a}$. That should clean things up and make it a better form to solve the problem.

$\endgroup$
0
$\begingroup$

This is one of my favourite tricks. Multiplying a series by a carefully chosen term. It is especially useful in dealing with arithmetic-geometric series. $$\begin{align} S &= a + 2a^2 + 3a^3 + \dots \\ aS &= a^2 + 2a^3 + 3 a^4 + \dots \\ S - aS &= a + a^2 + a^3 + \dots \\ S &= \frac {a + a^2 + a^3 + \dots}{1-a}\\ S &= \frac{a}{(1-a)^2}\\ \end{align}$$

The last step assumes $\left\lvert a \right\rvert \lt 1$. Otherwise, the infinite series does not converge.

$\endgroup$
  • $\begingroup$ I don't know why it isn't being rendered by MathJax $\endgroup$ – user230452 Feb 25 '16 at 5:08
  • $\begingroup$ @Anyone who knows Latex, please help and tell me what the mistake was $\endgroup$ – user230452 Feb 25 '16 at 5:12
  • $\begingroup$ first of all do not skip lines in between align (remove the two returns); second, use \\ instead of \ when you want to return to next line in align. $\endgroup$ – BGM Feb 25 '16 at 5:14
  • $\begingroup$ @BGM Thanks a lot. I'll make note of not to skip lines in align. $\endgroup$ – user230452 Feb 25 '16 at 5:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.