Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $f(x) = 1 + 2x + x^2 + 2x^3 +x^4+...$

If $c_{2n} = 1$ and $c_{2n+1} = 2$ $\forall n \ge 0$ find the interval of convergence and an explicit formula for $f(x)$.

The answers are $I = (-1,1)$ and $f(x) = \frac{1 + 2x}{1 - x^2}$

Can anyone please give me an idea how to get it...

Thanks in advance

share|cite|improve this question
Consider the series $1+x^2+(x^2)^2+\cdots$ and $2x(1+x^2+(x^2)^2+\cdots)$. – David Mitra May 23 '13 at 14:49
Note that, if the series converges absolutely, then you can rearrange the terms. – Mhenni Benghorbal May 23 '13 at 14:51
@DavidMitra: I just noticed your comment. You provide a third method of summation. You should put it in an answer (or I will add it to the two I wrote up, properly attributed, of course :-). Hmmm... I just noticed that yours is similar to Mhenni Benghorbal's sum, so I won't add it to mine. – robjohn May 23 '13 at 21:39
up vote 1 down vote accepted

The radius of convergence is $\dfrac1{\limsup\limits_{n\to\infty}|a_n|^{1/n}}$ and since $1\le a_n\le2$, the Squeeze Theorem says the radius of convergence is $1$.

$$ \begin{align} &\hphantom{(}1+2x\hphantom{)}+\hphantom{(}x^2+2x^3\hphantom{)}+\hphantom{(}x^4+2x^5\hphantom{)}+\hphantom{(}x^6+2x^7\hphantom{)}+\dots\\[8pt] =&(1+2x)+(1+2x)x^2+(1+2x)x^4+(1+2x)x^6+\dots\\[8pt] =&(1+2x)(1+x^2+x^4+x^6+\dots)\\[4pt] =&\frac{1+2x}{1-x^2} \end{align} $$

$$ \begin{align} &1+2x+x^2+2x^3+x^4+2x^5+x^6+2x^7+\dots\\[12pt] =&1+\hphantom{2}x+x^2+\hphantom{2}x^3+x^4+\hphantom{2}x^5+x^6+\hphantom{2}x^7+\dots\\ &\hphantom{1}+\hphantom{2}x\hphantom{\,+\;x^2}+\phantom{2}x^3\hphantom{\,+\;x^4}+\phantom{2}x^5\hphantom{\,+\;x^4}+\phantom{2}x^7\dots\\[4pt] =&\frac1{1-x}+\frac{x}{1-x^2}\\[9pt] =&\frac{1+2x}{1-x^2} \end{align} $$

share|cite|improve this answer
Thanks robjohn! – user78723 May 23 '13 at 23:29
Just throwing in a couple of other ways to look at the sum for consideration :-) – robjohn May 23 '13 at 23:31

Hint: Use the geometric series

$$ \sum_{m=0}^{\infty}z^m=\frac{1}{1-z}. $$

Added: Following my comment, first, we need to prove that the series converges absolutely. Now, notice this

$$ 1 + 2|x| + |x^2| + 2|x^3| +|x^4|+...\leq 2 + 2|x| + 2|x^2| + 2|x^3| +2|x^4|+... $$

$$ = 2\sum_{k=0}^{\infty} |x^k| = 2\sum_{k=0}^{\infty} |x|^k = \frac{1}{1-|x|},\quad |x|<1. $$

Thus the series converges absolutely. Now, you ca rearrange the series as

$$ f(x)= \sum_{k=0}^{\infty}x^{2k}+ 2\sum_{k=0}^{\infty}x^{2k+1} .$$

Use the hint and you should be able to finish the problem.

share|cite|improve this answer
Thanks Mhenni!. – user78723 May 23 '13 at 23:32
@user78723: You are welcome. – Mhenni Benghorbal May 24 '13 at 2:23

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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