Question

Would this series converge or diverge?

sum from n=1 to infinity (1/(3n-2)^(n+1/2) )? What test would you use?

Answer 1

\[\sum_{n=1}^{\infty} \frac{ 1 }{ (3n-2)^{n+\frac{ 1 }{ 2 }} }\]

Answer 2

have you tried a ratio test yet?

Answer 3

I didn't know what to do when I got
\[\lim_{n \rightarrow \infty} \ \frac{ (3n-2)^{n+\frac{ 1 }{ 2 }} }{ (3n+1)^{n+\frac{ 3 }{ 2 }} }\]
I'm thinking test for divergence where I treat this as a function, so I can just say it diverges?

Answer 4

I think a comparison test might get you somewhere. I'll try it out.

Answer 5

Looks like it does. Try comparing to
\[\sum_{n=1}^\infty\frac{1}{(3n-2)^n}\]
And use the root test to show convergence/divergence of the comparison series.

Answer 6

Root test? lol our professor skipped that section, we only went over ratio.

Answer 7

I'm pretty sure the ratio test also works.

Answer 8

integral test seem to be tricky at best ....

Answer 9

By the ratio test, you get a similar-looking limit:
\[\lim_{n\to\infty}\frac{(3n-2)^n}{(3n+1)^{n+1}}\\
\lim_{n\to\infty}\frac{1}{3n+1}\left(\frac{3n-2}{3n+1}\right)^n\\
\]

Answer 10

\[\sum_{n=1}^{\infty} \frac{ 1 }{ (3n-2)^{n+\frac{ 1 }{ 2 }} }\]
\[\sum_{n=1}^{\infty} \frac{ 1 }{ \sqrt{3n-2} }\frac{ 1 }{ (3n-2)^{n} }\]

Answer 11

^^^If you can figure out that limit, you can skip the comparison stuff.

Answer 12

the wolf says it converges .... :)

Answer 13

am I to use the ratio test on the function you separated? \[\lim_{n \rightarrow \infty} \frac{ (3n-2)^{1/2} }{ (3n+1)^{1/2} }\frac{ (3n-2)^{n} }{ (3n+3)^{n+1} }\]
looks messy?

Answer 14

\[\frac{ \sqrt{3n-2} }{ \sqrt{3n+1} }\frac{ (3n-2)^{n} }{ (3n+1)^{n+1} }\]
\[\cancel{\sqrt{\frac{ {3n-2} }{ {3n+1}}}}^1\frac{ (3n-2)^{n} }{ (3n+1)^{n+1} }\]
\[\cancel{\frac{ (3n)^n+....}{ (3n)^{n+1}+.... }}^0\text{ since the bottom is a higher degree than the top}\]

Answer 15

so id say it does converge

Answer 16

That's brilliant :) Thanks

Answer 17

.. good luck ;)