Question

Help! Interval of convergence!

Answer 1

okay so have you used the ratio test?

Answer 2

well since you ain't use the ratio test and set the remaining | value | < 1
and that gives you your range once you have that plug in each bound to the original formula and I got it to diverge both times so I think its (-1,1)

Answer 3

\[\lim_{n \rightarrow \infty} \left| \frac{ x ^{n+1}\ln(n+1) }{ n+1 }\times \frac{ n }{ x^n \ln n }\right|\]
\[\left| x \right|\lim_{n \rightarrow \infty} \left| \frac{ n }{ n+1 }\times \frac{ \ln (n+1) }{ \ln n }\right|\]
\[\left| x \right| <1 \]
\[-1<x<1\]

Answer 4

for x = -1
\[\lim_{n \rightarrow \infty}\frac{ (-1^n)\ln n(-1)^n }{ n } = \lim_{n \rightarrow \infty}\frac{ \ln n }{ n }\] what test should i use to prove its convergence or divergence?

Answer 5

for x=1
\[\lim_{n \rightarrow \infty}\frac{ (-1)^n \ln n }{ n }\] using alternating series, the series converges.. im not sure how you got diverging to both?

Answer 6

I got double divergence because it was 4 am lol. I think for the first one you'd use comparison to the 1/n harmonic function. I'm just assuming you know that.
and the second one does converge my b

Answer 7

okay thank you! :D