Question

Determine whether the following series converges or diverges.

Answer 1

First quick check before using any criterion: are the terms becoming small ?

Answer 2

the terms are not becoming small

Answer 3

I'd agree. (k+1)/k approaches 1 as k becomes larger and larger, but never reaches 1. And that exponent, 3k^2, increases fast. consider that you are adding an infinite number of terms together, all slightly greater than 1. Convergent? Divergent?

There are other ways to look at this matter.

You could let y=[ (k+1)/k ]^(3k^2) and then take the log of both sides. With some manipulation you may be able to apply l'Hopital's Rule to demo, analytically, whether or not the series converges.

Answer 4

"every term is greater than 1".

Answer 5

@mathmale how would that work? the second way, taking the log of both sides.

Answer 6

It "works" by allowing us to apply l'Hopital's rule. Have you studied that rule?

Answer 7

You can try, but noticing that every term is\(t_k > 1\) is enough:
\(\sum_{k=2}^\infty t_k \ge \sum_{k=2}^\infty 1 = \infty \).

Answer 8

Use the Divergence test?


re-write as \(( 1 + \frac{1}{k} )^{3k^2}\)

\(p = 3 k^2 \) so you have

\(\lim\limits_{p \to \infty} (1 + \dfrac{1}{\sqrt{\frac{p}{3}}})^p\)

and binomial expansion....

.....ripped from wiki: \( \lim_{n \to \infty}a_n \ne 0\), then the series must diverge