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

In order to study the series $\sum u_n$ where

$$u_n=\prod_{k=2}^n \left(1+\frac{(-1)^k}{\sqrt{k}} \right), $$

I'm trying to express $$\ln u_n= \sum_{k=2}^n \ln \left(1+\frac{(-1)^k}{\sqrt{k}} \right) $$

with asymptotic terms. I can write $$ \ln \left(1+\frac{(-1)^n}{\sqrt{n}} \right)=\frac{(-1)^n}{\sqrt{n}}-\frac{1}{2n}+\frac{(-1)^n}{3n^{3/2}}+ o\left(\frac {1}{n^{3/2}} \right) $$

but what about the previous terms of the sum? Can I do the same for each of those terms?

$$ n \geq N $$

$$ \ln(u_n)=\ln(u_N)+\ln(v_n) $$

$$ \ln(v_n)=\sum_{k=N+1}^n \ln (1+\frac{(-1)^k}{\sqrt{k}})=\sum_{k=N+1}^n \frac{(-1)^k}{\sqrt{k}}-\frac{1}{2k}+O(\frac{1}{n^{3/2}}) $$

$$ u_n=u_N\exp(\sum_{k=N+1}^n \frac{(-1)^k}{\sqrt{k}}-\frac{1}{2k}+O(\frac{1}{n^{3/2}})) $$

$$ u_n=u_N(1+\sum_{k=N+1}^n \frac{(-1)^k}{\sqrt{k}}-\frac{1}{2k}+\frac{1}{2}(\sum_{k=N+1}^n \frac{1}{k}+2\sum_{N+1\leq p,q\leq n, p\neq q} \frac{(-1)^{p+q}}{\sqrt{pq}}+O(\frac{1}{n^{3/2}})))$$

$$ u_n=u_N(1+\sum_{k=N+1}^n \frac{(-1)^k}{\sqrt{k}}+\sum_{N+1\leq p,q\leq n, p\neq q} \frac{(-1)^{p+q}}{\sqrt{pq}}+O(\frac{1}{n^{3/2}})))$$


share|cite|improve this question
Please do not use $$...$$ within the title. Also it is recommended that the title is not just math but contains some text as well, so I edited the title. And, exactly what do you want to do with the series? (That is, what does "study a series" mean?) Determine if it is convergent or divergent? Understand its asymptotic growth? – Srivatsan Oct 29 '11 at 16:03
Pick $N$ large enough and write every $u_n$ with $n\geqslant N$ as $u_n=u_Nv_n$. Then $u_N\ne0$ and your estimates apply to $(v_n)_{n\geqslant N}$, hence you are done. – Did Oct 29 '11 at 16:03
up vote 4 down vote accepted

Once you decided that you wanted to prove that the series $\sum\limits_nu_n$ diverges (you did, didn't you?), a simple approach is to try to bound $u_n$ from below. To this end, note that $\sqrt{2k+1}\geqslant\sqrt{2k}$ for every $k\geqslant1$ hence $$ u_{2n+1}=\prod_{k=1}^n\left(1+\frac1{\sqrt{2k}}\right)\cdot\left(1-\frac1{\sqrt{2k+1}}\right) $$ yields $$ u_{2n+1}\geqslant\prod_{k=1}^n\left(1+\frac1{\sqrt{2k}}\right)\cdot\left(1-\frac1{\sqrt{2k}}\right)=\prod_{k=1}^n\left(1-\frac1{2k}\right), $$ for every $n\geqslant1$. Again with the idea to use the simple lower bounds, note that $(1-\frac12x)^2\geqslant1-x$ for every $x$, hence $$ u_{2n+1}^2\geqslant\frac14\prod_{k=2}^n\left(1-\frac1{2k}\right)^2\geqslant\frac14\prod_{k=2}^n\left(1-\frac1{k}\right)=\frac1{4n}. $$ This proves that $u_{2n+1}\geqslant\frac12\frac1{\sqrt{n}}$. Since $u_{2n}\geqslant u_{2n+1}$, $u_{2n}+u_{2n+1}\geqslant\frac1{\sqrt{n}}$ for every $n\geqslant1$. Thus, $$ \sum\limits_{k=2}^{2n+1}u_k\geqslant\sum\limits_{k=1}^n\frac1{\sqrt{k}}=\int\limits_1^{n+1}\frac{\mathrm dt}{\sqrt{\lfloor t\rfloor}}\geqslant\int\limits_1^{n+1}\frac{\mathrm dt}{\sqrt{t}}=2\sqrt{n+1}-2. $$ This proves that the sequence of general term $\sum\limits_{k=2}^{n}u_k$ is unbounded hence the series $\sum\limits_nu_n$ diverges.

share|cite|improve this answer

Is it normal that I cannot comment questions (I can comment my own answers) if I'm a new user?

To the point: You can say that you don't care about $k$s smaller than some $K$: You can always skip the elements of the sum with $n$ smaller than $K$ and it will not change the fact if the series converges or not. Then, if you skip factors with $k<K$ in every element of the (shortened) sum, it also won't change the convergence (it is just a non-zero constant). If $K$ is big enough ($k \ge K$) you can say that $|o(1/k^{3/2})|$ is smaller than $1/3k^{3/2}$ or something like that (moreover you can say that it is negative, but it probably will not help at all).

share|cite|improve this answer
Yeah, @Didier told exactly the same. You don't need $n^{-3/2}$, $o(1/n)$ is enough. I think you can finish it after our remarks. If you still need some hints, ask. – savick01 Oct 29 '11 at 16:36
"Is it normal that I cannot comment questions (I can comment my own answers) if I'm a new user?" - yes. Earn 50 rep first, and then you'll be able to make comments. – J. M. Oct 29 '11 at 16:47
I would like to know whether what I wrote is correct for I can't see why the series is divergent from my last equation. – Chon Oct 30 '11 at 20:46

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.