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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

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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
1  
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

2 Answers 2

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 $1$.


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


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

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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.

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Thanks Mhenni!. –  user78723 May 23 '13 at 23:32
    
@user78723: You are welcome. –  Mhenni Benghorbal May 24 '13 at 2:23

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