# Studying the series $\sum \prod\limits_{k=2}^n \left(1+\frac{(-1)^k}{\sqrt{k}} \right)$

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?

$$n \geq N$$

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

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

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

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

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

?

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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

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.
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).
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