Can anyone explain how to do this problem? I think you might be able to approach it with the ration test but I'm unsure. Any help is greatly appreciated!
$$\sum_{n=0}^{\infty} \frac{(2x-3)^n}{n \ \ln n}$$
Find the interval of convergence.
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2 Answers
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This sum needs to be from $n=2$ to $\infty$, because otherwise the first 2 terms aren't defined. Assuming this:
If we use the Ratio Test, we get $$\lim_{n\to\infty} \left|\frac{(2x-3)^{n+1}}{(n+1)\ln(n+1)}\cdot \frac{n\ln(n)}{(2x-3)^n}\right| = |2x-3| \lim_{n\to\infty} \left|\frac{n\ln(n)}{(n+1)\ln(n+1)}\right| = |2x-3|<1$$ which leads us to conclude precisely that the power series converges on
$-1 < 2x-3 < 1$, or $2 < 2x < 4$, or $1<x<2$.
Check the endpoints:
At $x=2$, we have $$\sum_{n=2}^\infty \frac{1}{n\ln(n)}$$ which diverges (check this; use the integral test).
At $x=1$, we have $$\sum_{n=2}^\infty \frac{(-1)^n}{n\ln(n)}$$ which converges by the alternating series test (check this).
The series thus converges on the interval $[1,2)$, converges absolutely on $(1,2)$, and has radius of convergence $R=\frac{1}{2}$.
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$\begingroup$ Thanks for the help! Are you sure there's no way to do it with n=0? My professor was specific with n=0 for that problem, but she very well may have made a mistake. $\endgroup$ Commented Apr 21, 2014 at 3:22
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$\begingroup$ The whole issue with starting with $n=0$ is $\ln(0) = ?$, $\frac{1}{1\cdot\ln(1)} = ?$. Your professor may have been specific about $n=0$, but it's relatively easy to make a mistake like that when writing up questions, particularly when it's something like the denominator vanishing or becoming undefined for one reason or another. If anything, it was probably a typo, and wasn't intended to be a focus of the problem. $\endgroup$ Commented Apr 21, 2014 at 13:22
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Using ratio test: $\displaystyle R = \lim_{n \to \infty} |\dfrac{a_{n+1}}{a_n}| = \lim_{n \to \infty} \left(|2x - 3|\times \dfrac{nlnn}{(n+1)ln(n+1)}\right) = |2x - 3|$. So the series converges if $R < 1 \iff |2x - 3| < 1 \iff -1 < 2x - 3 < 1 \iff 2 < 2x < 4 \iff 1 < x < 2$. So the interval of convergence is: $(1, 2)$ Note: please check the endpoints.
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$\begingroup$ It's not true that the series converges if and only if $|2x-3| < 1$ (it is true that the series converges if $|2x-3|<1$, but the converse is not in general true). The Ratio Test is inconclusive in the case $|2x-3| = 1$, so you need to handle the endpoints separately. $\endgroup$ Commented Apr 21, 2014 at 2:57
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$\begingroup$ Also, if $L = \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right|$, it's not even true that $\sum_{n=0}^\infty a_n x^n$ converges absolutely if and only if $L < 1$. The series $\sum_{n=1}^\infty \frac{x^n}{n^2}$ converges absolutely on $[-1,1]$, but the Ratio Test only tells you the series converges on $(-1,1)$ and leaves the endpoints as uncertain. An example where the Ratio Test does tell you the whole story is $\sum_{n=0}^\infty x^n$, in which case the Ratio Test says that this series converges on $(-1,1)$, which is also the interval where the series converges absolutely. $\endgroup$ Commented Apr 21, 2014 at 3:01
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$\begingroup$ @NicholasStull: I knew the endpoints, but didn't have time to type it up and post it.... $\endgroup$– DeepSeaCommented Apr 21, 2014 at 3:05
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$\begingroup$ Okay. My qualms were more with the implied equivalence. I figured this was likely an oversight, but wanted to point out examples for the OP's benefit. $\endgroup$ Commented Apr 21, 2014 at 3:08