$$\sum_{k=1}^{\infty} \frac{kx^{k+1}}{3^k}$$

Im guessing its equivalent to:

$$\sum_{k=1}^{\infty} k\left(\frac{-1}{3}\right)^k x^{k+1}$$

But I am not sure on how to advance past this step.

How would I find the function it represents?

Also do I need to find its interval of convergence and find the function over this interval or it doesn't necessarily matter?

  • $\begingroup$ I edited in LaTeX formatting, please check this, as it seems there is an error in the original question. $\endgroup$ – nbubis Mar 28 '14 at 0:10
  • 1
    $\begingroup$ I don't see how the two could possibly be equivalent. $\endgroup$ – Jared Mar 28 '14 at 0:12
  • $\begingroup$ About what $x$ value is this series evaluated? $\endgroup$ – user122283 Mar 28 '14 at 0:15
  • $\begingroup$ @SanathDevalapurkar Thats my question below the bold text $\endgroup$ – Sad CRUD Developer Mar 28 '14 at 0:38

First you need to write it as a proper Taylor series, which means the exponent should just be $k$, so change indexes:

$$ k' = k + 1 \rightarrow k = k' - 1 \\ \sum_1^\infty \frac{kx^{k + 1}}{3^k} = \sum_2^\infty \frac{(k - 1)x^k}{3^{k - 1}} = \frac{1}{3}\sum_2^\infty \left(k - 1\right)\left(\frac{x}{3}\right)^k $$

As per Pedro's hint, this does almost look like a geometric series (i.e. $\frac{1}{1 - x} = \sum x^i$). The fact that the first two terms are missing suggests that this might be a second derivative (but it just might be a first derivative...integrate and see where it goes). I would integrate this power series and see if you can come up with an easier function to deal with. As for the radius of convergence, the ratio test should show you that this series converges when $|x| < 3$.

Also, don't fret if some terms are missing, you can always just add them in then subtract them off. For instance you could write:

$$ \frac{1}{3}\sum_2^\infty \left(k - 1\right)\left(\frac{x}{3}\right)^k = \frac{1}{3}\left(\left(\sum_0^\infty \left(k - 1\right)\left(\frac{x}{3}\right)^k\right) + 1\right) $$

It's $+1$ because you need to subtract $(0 - 1) = -1$ and then the $k = 1$ term already gives $0$ in the sum, so no need to subtract off $0$.

  • $\begingroup$ Why calculate the derivative of k? (k') $\endgroup$ – Sad CRUD Developer Mar 28 '14 at 0:40
  • $\begingroup$ $k$ is a constant. You don't differentiate or integrate $k$. You should try integrating the power series with respect to $x$. I don't actually know that will work, but it might. $\endgroup$ – Jared Mar 28 '14 at 0:44
  • $\begingroup$ Is finding the derivative of a series the usual approach of getting a function as an answer? (Pardon my ignorance on the subject) $\endgroup$ – Sad CRUD Developer Mar 28 '14 at 0:54
  • $\begingroup$ It's just a trick you should keep in your back pocket. If you do enough of these problems you'll be able to recognize when this is likely to work and when it isn't. $\endgroup$ – Jared Mar 28 '14 at 0:56

Hint It is of the form $x\sum k(\alpha x)^k$, and the derivative of $\sum x^k$ is $\sum k x^{k-1}$.


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