Question

Does the following converge or diverge?

Answer 1

hi

Answer 2

\[\sum_{k=1}^{\infty} \frac{3k}{4^{3k-1}}\] I'm not sure how to go about simplifying this thing. I'm not sure if I can do the Basic Divergence Test on this equation as it is.

Answer 3

you can do the ratio test

\(\color{black}{\large \displaystyle \lim_{k\to\infty}\left[\frac{3(k+1)}{4^{3(k+1)-1}}\times \frac{4^{3k-1}}{3k}\right]=\lim_{k\to\infty}\left[\frac{3(k+1)}{4^{3k+2}}\times \frac{4^{3k-1}}{3k}\right]=\frac{1}{4^{3}}<1}\)

Answer 4

you can also use the comparison test that \[\sum_{k=1}^{\infty} \frac{3k}{4^{3k-1}}=(3/4) \sum_{k=1}^{\infty} \frac{k}{64^{k}}<(3/4) \sum_{k=1}^{\infty} \frac{k}{k^3}\]

Answer 5

Thank ye.