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Find the interval of convergence of the power series. Be sure to include a check for convergence at the endpoints of the interval. $$(a) \ \sum_{n=1}^\infty \frac{(-1)^n x^n}{n} \qquad (b) \ \sum_{n=1}^\infty \frac{(-1)^{n+1}(x-5)^n}{n\cdot 5^n}$$

I know you use the power series and start with ratio test , test the points at the end. Here is my work so far. Does it look like I am doing the right thing for B? (The solution for A is in an answer)

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  • $\begingroup$ So far i tried the ratio test. I posted my results $\endgroup$ – Zak Dec 11 '14 at 0:27
  • $\begingroup$ Does it look like I am doing the right thing for B? Thanks $\endgroup$ – Zak Dec 11 '14 at 1:32
  • $\begingroup$ When you factor $-1/5$ out of absolute values, you should be taking the absolute value of this, making it $\frac{1}{5}|x-5|<1$. $\endgroup$ – Addem Dec 11 '14 at 2:14
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At this point you can factor $x$ from your expression to obtain

$\displaystyle |x| \lim_{n\rightarrow \infty}\left| \frac{n}{n+1}\right|$. You should be able to find this limiting value.

At that point, you know that the series converges when this expression is less than 1.

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I believe this is the answer for the first problem. Please correct me if I am wrong

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  • $\begingroup$ Your solution for (a) is nearly correct. You correctly compute the limit in the Ratio Test and give the correct bounds on $x$. For when $x=-1$ you don't get an alternating series, though, since $(-1)^{n}\cdot (-1)^{n} = (-1)^{2n} = 1$ for all $n$. So this is just the harmonic series which doesn't converge. It is the alternating series you give when $x=1$ though, so it converges there. $\endgroup$ – Addem Dec 11 '14 at 2:12

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