# Thread: [SOLVED] Series - Converge or diverge

1. ## [SOLVED] Series - Converge or diverge

Hello,

How can I show if these series are convergent or divergent?

k = 1∞ cos((k^(3))/(3^(k))

k = 1∞ ((-1)^(k))((sin(1/k))/(k))

Appriciate any guidance and a detail explanation.

Thank you

2. recall $|\cos x|\le1,$ then first series converges absolutely by direct comparison test by the geometric series.

as for the second one, try Leibniz test.

3. Originally Posted by Krizalid
recall $|\cos x|\le1,$ then first series converges absolutely by direct comparison test by the geometric series.

as for the second one, try Leibniz test.
Hmm...

I thought the first one diverges...

Doesn't $\lim_{k \to \infty} \frac{k^3}{3^k} = 0$ which would mean the sum just becomes 1+1+1+1+1+... (or near enough 1)...

I just checked in Maple and you have...

$\sum_{k=1}^{n} \cos(\tfrac{3^k}{k^3})$ tends to n-1 (or n-1.3 or something like that...) for n large

4. The first one is pretty easy.

You've got:

$
\lim_{k \to \infty} cos(\frac{k^3}{3^k}) =cos( \lim_{k \to \infty} \frac{k^3}{3^k})=cos0=1 \neq 0$

and it is divergent. If it was 0 the test would've been inconclusive. (limit test)

For the second one:

you can see that $\lim_{x->\infty}\frac{sin(\frac{1}{x})}{x}=0$

and it is decreasing (by drawing the graphic of the function $\frac{sin(\frac{1}{x})}{x}$.

We can conclude that it is alternating series which converge.

Regards.

5. Originally Posted by Krizalid
recall $|\cos x|\le1,$ then first series converges absolutely by direct comparison test by the geometric series.

as for the second one, try Leibniz test.
I think the OP means $\cos \left( \frac{k^3}{3^k} \right)$ (but it is very easy to misread it, even when reading it carefully).

6. yes, i misread it, that's why my answer doesn't fit here.