1. ## Series question

I am having trouble solving a problem involving a series.

The problem is the following:

$x \displaystyle\sum_{n=1}^{\infty}\ na_n x^{n-1} + \displaystyle\sum_{k=0}^{\infty}\ a_k x^k$

The problem wants me to rewrite the given expression as a sum who generic term involves $x^n$

For some reason I can't get the same answer that is in the back of the book. I especially don't know what to do when I have the variable "n" with the variable "k". I've tried substituting (n+1) in the place of "n" but that still does work. That method worked in similar problems, so I feel I'm missing out on a subsequent step.

Can anyone please explain thi s to me?

Thank you!

2. ## Two Hints

First, what happens when you use the distributive property on the product $x\cdot\sum_{n=1}^{\infty}na_nx^{n-1}$?

Second, what is the value of $na_nx^n$ when n=0?

--Kevin C.

3. ## didnot understand

Originally Posted by TheBerkeleyBoss
I am having trouble solving a problem involving a series.

The problem is the following:

$x \displaystyle\sum_{n=1}^{\infty}\ na_n x^{n-1} + \displaystyle\sum_{k=0}^{\infty}\ a_k x^k$

The problem wants me to rewrite the given expression as a sum who generic term involves $x^n$

For some reason I can't get the same answer that is in the back of the book. I especially don't know what to do when I have the variable "n" with the variable "k". I've tried substituting (n+1) in the place of "n" but that still does work. That method worked in similar problems, so I feel I'm missing out on a subsequent step.

Can anyone please explain thi s to me?

Thank you!
i didnot get you. first what is meant by generic term? it would be helpful if you could cite a simple example of a similar problem.

4. If I factor the "x" in using the distributive property, I get the following result:

$\displaystyle\sum_{n=0}^{\infty}\ na_n x^{n}$ with the index of summation set equal to zero, correct? If n=0 in the above variables, then the whole thing is equal to zero?

I'm still confused. I know the answer is supposed to be $(n+1)a_n x^{n}$. I can get everything to look right by using n = (n+1) except for the $a_n$ part. I figured it would be $a_{n+1}$ but it is not. I'm guessing the $a_k x^k$ has something to do with it.

5. My first hint was that $x\cdot\sum_{n=1}^{\infty}na_nx^{n-1}=\sum_{n=1}^{\infty}x\cdot{n}a_nx^{n-1}=\sum_{n=1}^{\infty}na_nx^n$.

The second was to note that
$\sum_{n=0}^{\infty}na_nx^n=0a_0+1a_1x+2a_2x^2+\dot s$
$=0+1a_1x+2a_2x^2+\dots$
$=1a_1x+2a_2x^2+\dots$
$\sum_{n=0}^{\infty}na_nx^n=\sum_{n=1}^{\infty}na_n x^n$
Since the n=0 term (not the entire sum, just that one term) is zero, we can add it to the series without changing the value of the series.
Thus, we go from
$x\sum_{n=1}^{\infty}na_nx^{n-1}+\sum_{k=0}^{\infty}a_kx^k$
to
$\sum_{n=1}^{\infty}na_nx^n+\sum_{k=0}^{\infty}a_kx ^k$
to
$\sum_{n=0}^{\infty}na_nx^n+\sum_{k=0}^{\infty}a_kx ^k$
Now, just rename the variable "k" to "n"
$\sum_{n=0}^{\infty}na_nx^n+\sum_{n=0}^{\infty}a_nx ^n$,
and combine the sums, grouping terms of the same n
$\sum_{n=0}^{\infty}\left(na_nx^n+a_nx^n\right)$,
$=\sum_{n=0}^{\infty}(n+1)a_nx^n$.
Does that make sense?

--Kevin C.