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We have a sum $\sum_{0}^{\infty} 2^{n+1}(x+1)^{3n+1}$ I am asked to find all values of x s.t this sum converges then compute the sum.

I have no idea what im doing so i did this. $\frac {2^{n}(x+1)^{3n}} {2^{n+1}(x+1)^{3n+1}}$

this yields $\frac {1}{2(x+1)^{3}}$ this looks alittle wierd but i just move it over and solved to get $x=-1 +2^{-1/3}$ this isnt right i also have no idea to find the sum if it was please help.

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3 Answers 3

up vote 2 down vote accepted

First you should realize that just about the only series you've learned how to get an explicit sum for is a geometric series. So your first goal should be to make the given series looks like the standard form of a geometric series. One way to do this is as follows: \begin{align} \sum_{n = 0}^\infty 2^{n+1} (x+1)^{3n+1} & = 2(x+1) \sum_{n = 0}^\infty 2^{n} (x+1)^{3n} \\ & = 2(x+1) \sum_{n = 0}^\infty (2(x+1)^3)^n. \end{align} Now $\displaystyle \sum_{n = 0}^\infty (2(x+1)^3)^n$ is a geometric series with $a = 1$ and $r = 2(x+1)^3$. Hence $$\sum_{n = 0}^\infty (2(x+1)^3)^n = \frac{1}{1 - 2(x+1)^3},$$ and from the work about it follows that $$\sum_{n = 0}^\infty 2^{n+1} (x+1)^{3n+1} = 2(x+1) \sum_{n = 0}^\infty (2(x+1)^3)^n = \frac{2(x + 1)}{1 - 2(x+1)^3}.$$

A geometric series $\displaystyle \sum_{n=0}^\infty ar^n$ converges if and only if $|r| < 1$, so our given series converges if and only if $$|2(x+1)^3| < 1 \iff -1 - 2^{-1/3} < x < -1 + 2^{-1/3}.$$

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that is really obvious i just couldn't understand what to make a O.o thank you. –  Faust7 May 10 '13 at 22:35
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Recall the geometric series $$a+ar+ar^2+ar^3+\cdots+ar^n+\cdots \tag{$\star$}$$ converges for $\vert r \vert < 1$ and equals $\dfrac{a}{1-r}$.

Try to write your series in the above form $(\star)$, for a suitable $a$ and $r$.

EDIT

$$\sum_{n=0}^{\infty} 2^{n+1} (x+1)^{3n+1} = 2(x+1) \sum_{n=0}^{\infty}\left(2(x+1)^3\right)^n$$ Can you now see it?

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well tis nothing like that form but r or is 2(x+1) the other is $2(x+1)^{3}$ –  Faust7 May 10 '13 at 22:30
    
@Faust7 I added some more to the answer. Can you now see it? –  user17762 May 10 '13 at 22:32
    
not really i wrote that down on my paper before i tried the gibberish above that exact line and had no idea what to do with it so i tried something else. –  Faust7 May 10 '13 at 22:33
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Your original sum is:

$$\sum_{n=0}^\infty{2^{n+1}(x+1)^{3n+1}}$$

We can rearrange the summation as

$$\sum_{n=0}^\infty{2 \cdot (2^n)(x+1)^{3n}(x+1)}$$ $$=\sum_{n=0}^\infty{2\left(2(x+1)^3\right)^n(x+1)}$$ $$=\sum_{n=0}^\infty{\left(2(x+1)^3\right)^n2(x+1)}$$

So we are interested in what happens when $n$ approaches infinity. In order for this sum to converge we need $2(x+1)^3$ to get smaller and smaller. Do you know what to do next?

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