# Homework Help: Series problems convergent or divergent

Tags:
1. Mar 10, 2017

### kevin3295

Poster warned that the homework template is not optional.
Determine if they are convergent or divergent, If it converges find the sum:

∑ 3^(n-1) 2^n
n=1

∑ ln(1/n)
n=1

∑ tan^n ( π/6)
n=1

I tried to find information on how to solve them but I couldn't, thanks for the help

Last edited by a moderator: Mar 10, 2017
2. Mar 10, 2017

### nuuskur

What is the necessary condition for convergence of a series? I.e if a series converges, then ...? By this ... condition, we immediately deduce that your second series diverges.

If I'm reading this right, the first series looks like
$$\sum_{n=1}^\infty 3^{n-1}\cdot 2^n = \sum_{n=1}^\infty \frac{6^n}{3},$$
which clearly diverges by ...

The third series looks to be
$$\sum_{n=1}^\infty \left (\tan \frac{\pi}{6}\right )^n = \sum_{n=1}^\infty \left (\frac{1}{\sqrt{3}}\right )^n$$

Last edited: Mar 10, 2017
3. Mar 10, 2017

### kevin3295

Thanks for answering, I am studying series on my own and I am still not sure how to solve them but I would like to use these three exercises as examples, it would be very helpful to have the procedure.

4. Mar 10, 2017

### Staff: Mentor

Are you using a textbook? Any textbook that covers series will have a number of theorems (tools) that you can use to decide whether a series converges or diverges. There's not a "cookbook" procedure, as such, but there are tests such as the Nth Term Test for Divergence, Comparison Test, Limit Comparison Test, Integral Test, Ratio Test, and Alternating Series Test.
It's possible that for a given series, more than one test will produce a result. It's also possible that one test does not tell you anything, so you try a different test. It should never happen that two tests produce opposite results.

A textbook will also present several simple series whose behavior is known, that can be used in the tests where you compare a series you're investigating with another series that is known to converge or known to converge.

5. Mar 10, 2017

### nuuskur

Let $\sum\limits_{k=1}^\infty a_k$ be a series, where for every $k\quad a_k\in\mathbb{R}$. An elementary result is the following necessary condition for convergence: If a series converges, then $\lim_{n\to\infty} a_n = 0$. Therefore, if a series fails to meet this condition, it diverges. For instance, $\lim_{n\to\infty}\ln \frac{1}{n}=-\infty$, therefore the second series diverges.

A necessary condition, however, need not be sufficient i.e if $\lim_{n\to\infty}a_n=0$ it's not necessary that the series converges. At this point we start using the Cauchy, d'Alembert tests and possibly something else, entirely. Perhaps none of the tests are conclusive and we need to prove convergence by definition. [the $\varepsilon - N$-language to be specific].

Last edited: Mar 10, 2017
6. Mar 10, 2017

### Staff: Mentor

@kevin3295, in future posts, please do not delete the homework template. Its use is required here.

7. Mar 11, 2017