0
$\begingroup$

I was working on this and thought that it didn't matter what my function was. However I noticed in order to solve it like I usually do. It has to look like $$\sum a_k x^k$$ However mine does not look like that. It does not have an $x^k$. $$\sum_{k=2}^\infty x^k (1-x)$$

I am positive the sum exists but don't know how I should find it.

$\endgroup$
3
  • $\begingroup$ Try splitting it up into two sums, using $x^k(1-x)=x^k-x^{k+1}$. $\endgroup$ Dec 7, 2017 at 22:29
  • $\begingroup$ I believe I have to show that they are both convergent in order to do that? So if I show they are both convergent when $|x| < 1$ then is it okay to use it? $\endgroup$
    – John.Doh
    Dec 7, 2017 at 22:35
  • $\begingroup$ Yes, this is valid if $|x|<1$. $\endgroup$ Dec 7, 2017 at 22:43

2 Answers 2

1
$\begingroup$

Note that $$\sum_{k=2}^{\infty}x^k(1-x)=x^2(1-x)\sum_{i=0}^{\infty}x^i =x^2,$$ if $|x|<1$.


If $|y|<1$ then $$\frac{1}{1-y} = 1+y+y^2+\cdots = \sum_{i=0}^{\infty} y^i.$$

$\endgroup$
1
$\begingroup$

Suppose $|x|<1$; then \begin{align*} \sum_{k=2}^{+\infty}x^k(1-x) &=(1-x)\sum_{k=2}^{+\infty}x^k\\ &=(1-x)\sum_{k=0}^{+\infty}x^k-(1-x^2)\\ &=\frac{1-x}{1-x}-1+x^2\\ &=x^2 \end{align*}

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .