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

I have the following series:

$\sum_n{a_n}$, where $a_1 = 1$, $a_{n+1} = a_n \,\frac{2 + \cos n}{\sqrt n}$. Does it converge or diverge?

Edit: formatted.

share|cite|improve this question
Please do use LaTeX to properly write mathematics. I tried to edit your post but it confuses me. In FAQ you can find some directions for this. – DonAntonio Nov 20 '12 at 5:17
@DonAntonio You got it right! I did not know it was using LaTeX at all. May you help me to figure it out now?> – cybertextron Nov 20 '12 at 5:19

Use the Ratio Test. You can read about it here.

share|cite|improve this answer
I already the Ratio test. I now need to know if cos(n)/sqrt(n) diverges or converges – cybertextron Nov 20 '12 at 5:26
As a sequence, it converges to $0$. The numerator is bounded and the denominator is unbounded and increasing. – alex.jordan Nov 20 '12 at 5:28
I applied the Ratio test: I got (2 + cos( n + 1 )) / (n + 1 ). As this limit goes to 0 when n-> infinite, then the series a_n also converges? – cybertextron Nov 20 '12 at 5:30
Yep. If the limit of the ratio of a term to the previous term is less than 1, then the terms can be summed to a finite sum. – alex.jordan Nov 20 '12 at 5:33

Consider the sequence $a_{n+1}=a_n\frac{3}{\sqrt{n}}$. See whether it converges or not. Now see if this converges, why should the original sequence converge.

share|cite|improve this answer
I got 3 / sqrt( n + 1 ) when I applied the ratio test for the sequence you gave me. Am I right? – cybertextron Nov 20 '12 at 5:34
You have to see the limit for the ratio test. What is the ratio $|\frac{a_{n+1}}{a_n}|$ at the limit $n$ tends to infinity. Now see that each term of the original sequence is upper bounded by each term in this sequence. – dineshdileep Nov 20 '12 at 5:38

Continuing with Alex's answer: apply now Dirichlet's test*, with the monotone descending sequence $\,\displaystyle{\left\{\frac{1}{\sqrt n}\right\}}\,$ and the bounded sequence $\,\displaystyle{\left\{\sum_{k=1}^n\cos n\right\}}\,$

share|cite|improve this answer

Alternatively, use direct comparison with a geometric series, noting that all terms are positive. The first 15 terms sum to whatever they sum to. After that, with $n\ge16$, then $a_{n+1}=a_n\frac{2+\cos(n)}{\sqrt{n}}\le\frac{3}{4}a_n$. Inductively, $a_n\le\left(\frac34\right)^{n-16}a_{16}$ for $n\geq16$. So the series sums to at most $$\sum_{n=1}^{15}a_n+a_{16}\sum_{n=16}^\infty\left(\frac34\right)^{n-16}$$ or rather $$\sum_{n=1}^{15}a_n+4a_{16}$$

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.