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I have found the radius of convergence $R=2$ and the interval of convergence $I =[-2,2)$ for the following infinite series:

$\sum_{n=1}^\infty (-1)^n \frac{ (x+2)^n }{n2^n}$

Approach:

let $a_n = (-1)^n \frac{ (x+2)^n }{n2^n}$

Take the ratio test

$\lim_{x\to -\infty} \left| \frac{a_{n+1}}{a_n} \right| = \left| -\frac{1}{2}(x+2)\right|$

So it converges for $-2 < (x+2) < 2, R = 2$

Test end points

if $x = -2$

$\sum_{n=1}^\infty (-1)^n(0) =0 < \infty$ , converges

if $x = 2$

$\sum_{n=1}^\infty (-1)^n \frac{4^n}{n2^n} = \sum_{n=1}^\infty (-1)^n \frac{2^{2n}}{n2^n} = \sum_{n=1}^\infty (-1)^n \frac{2^n}{n}$

let $b_n = \frac{2^n}{n}$

$b_{n+1} > b_n$ for every value of $n$

By the Alternating Series Test, $\sum_{n=1}^\infty (-1)^n b_n$ diverges

Hence, $R = 2$ and $I = [-2, 2)$

Is my approach correct? Are there any flaws, or better ways to solve?

theta
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1 Answers1

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The radius of convergence is right. The interval of convergence started out OK, you wrote that for sure we have convergence if $-2\lt x+2\lt 2$. But this should become $-4\lt x\lt 0$.

The endpoint testing was inevitably wrong, since the incorrect endpoints were being tested.

We have convergence at $x=0$ (alternating series) and divergence at $x=-4$ (harmonic series).

André Nicolas
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  • I do not know why my head interpreted x+2 as x*2 – theta Nov 08 '13 at 21:21
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    Just a minor slip. The endpoint testing should have alerted you. In the calculation at $x=2$, you got terms $\frac{2^n}{n}$, very badly divergent (the alternating series stuff you mentioned is really irrelevant). It is clear that we have divergence also much earlier. – André Nicolas Nov 08 '13 at 21:27