Question

find the interval of convergence of the series:
summation from n=2 to infinity of
(x^(8n))/ (n(ln n)^4)

Answer 1

I've found the radius of convergence to be 1 and my endpoints to be -11 and 11. my problem is i can't figure out what test to use to determine if their convergent or divergent at those endpoints

Answer 2

How are your end points -11 and 11 if the radius of convergence is 1?

Answer 3

\[\sum_{2}^{\infty} \frac{ x ^{8n} }{ n(\ln(n)^4 }\]

Answer 4

sorry! i meant the radius of convergence is 1 and my endpoints are -1 and 1

Answer 5

@kimmy0394 consider each case separately. For \(x=1\), recognize that since \(1^{8n}=1\) our series reduces to \(\sum_{n=2}^\infty\frac1{n(\log n)^4}\). Notice we have \(\dfrac1n\) and \(\log n\), which are related since \(\int\frac1x=\log x\). Hmmm... perhaps we could use the *integral* test? We check our integral for convergence as follows. $$\begin{align*}\int_2^\infty\frac1x\frac1{\log^4x}\mathrm{d}x&=\int_{\log2}^\infty\frac1{u^4}\mathrm{d}u\text{ where }u=\log x\\&=\lim_{c\to\infty}\int_{\log2}^cu^{-4}\mathrm{d}u\\&=\lim_{c\to\infty}\left[-3u^{-3}\right]_{\log2}^c\\&=3\lim_{c\to\infty}\left(\frac1{\log^32}-\frac1{c^3}\right)\\&=\frac3{\log^32}\end{align*}$$Since this integral converges, so does our series by the integral test.
(note you need to show that our series terms are decreasing and positive to justify the integral test)

Answer 6

Then for \(x=-1\), recognize that \((-1)^{8n}=\left((-1)^8\right)^n=1^n=1\) and so our series reduces in an identical manner. Thus since our series converges at \(x=1\) it also converges at \(x=-1\). Overall, our inteval of convergence is then \([-1,1]\).