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Find values of $x$ for which the series converges and compute that sum.

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

I'm not even really sure how to approach this problem, I have several theorems in my textbook but I am not used to tackling something like this unless via induction instead of using calculus.

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With a little bit of algebra, you can bring this back to a geometric series. –  1015 Mar 31 '13 at 23:08
    
It's a geometric series. I'd start by looking up that topic in the calc book (usually covered in calc 2, sometimes introduced in calc 1). –  coffeemath Mar 31 '13 at 23:09
    
Use ratio test to prove the convergence. –  Mhenni Benghorbal Apr 1 '13 at 2:43
    
can i buy a 0 anyone? –  Faust7 Apr 1 '13 at 17:03

1 Answer 1

Hints:

$$(1)\;\;\;\;2^{n+1}(x+1)^{3n+1}=2(x+1)\left[2(x+1)^3\right]^n$$

$$(2)\;\;\;\;\left|2(x+1)^3\right|<1\iff|x+1|<\frac{1}{\sqrt[3]2}\;\ldots$$

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