Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The sequence $a_k$ = $$3/2\sum_{k=1}^{\infty}\frac{1}{4^k}$$

Does the series converge? Compute $\liminf a_k^{1/k}$, $\limsup a_k^{1/k}$, $\liminf a_{k+1}/a_k$ and $\limsup a_{k+1}/a_k$ as $k\to\infty$

How do we show this? Thank you!

I think by ratio test, this is convergent. And $a_{k+1}/a_k$ = 1/4 and then?

share|cite|improve this question
This question is a mess. Do you really mean $a_k=3/2$? What does that have to do with the series? – Thomas Andrews Dec 27 '12 at 15:13
Why do you show NOTHING about your thoughts on this (awfully badly formulated) question? The less you put into the system, the less you should expect to receive from it... – Did Dec 27 '12 at 15:14
So you really mean $a_k=\frac{3}{2}\frac{1}{4^k}$? – Thomas Andrews Dec 27 '12 at 15:18
up vote 2 down vote accepted

As written, your definition of the $a_k$ terms makes no sense. It's possible that you were indicating a constant sequence, with each term $$a_k:=\frac32\sum_{j=1}^\infty\frac1{4^j},$$ but the fact that you're asking whether "the series" converges suggests that this is not what you meant. I suspect that your sequence is instead given by $$a_k:=\frac32\cdot\frac1{4^k}$$ for all $k$, at which point you'd be interested in knowing if the series $\sum_{k=1}^\infty a_k$ converges, and the rest of the problem amounts to using the root and ratio tests to determine this. You've correctly used the ratio test to confirm that the series converges. All that's left is to perform the root test. To do this, you'll want to observe (and possibly prove first) that for any positive real number $\alpha$, we have $$1=\lim_{k\to\infty}\alpha^{1/k}.$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.