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Is the series $$ \sum u_{n}$$

$$ u_{n}=n!\prod_{k=1}^n \sin\left(\frac{x}{k}\right)$$

$$ x\in]0,\pi/2] $$

convergent or divergent?

We have:

$$ u_{n}\leq n!\prod_{k=1}^n \frac{x}{k}$$

$$ u_{n}\leq x^n$$

If $0<x<1$ the series is convergent.

$$ u_{n}\geq n! \prod_{k=1}^n \frac{2x}{\pi k}$$

$$ u_{n} \geq \prod_{k=1}^n \frac{2x}{\pi}$$

If $x=\pi/2$, $u_{n}\geq1$, $\sum u_{n}$ is divergent.

What about the case $x\in[1,\pi/2[$ ?

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As N.S. shows in his answer, the ratio test quickly resolves the problem for all $x\neq 1$, so for now fix $x=1.$ Then $\displaystyle u_n = \prod_{k=1}^n k\sin(1/k).$

We show that $\displaystyle \lim_{n\to\infty} u_n = L> 0 $ so $\sum u_n$ diverges.

Since $\sin x<x$ we see $u_n<1.$ In order to apply a convergence criterion for infinite products, let us work with $v_n = 1/u_n$ instead. Let us remember that "convergence" of an infinite product demands the partial products tends to a finite non-zero limit. We compute $$v_n=\prod_{k=1}^n \frac{1}{k\sin(1/k)} = \prod_{k=1}^n (1+a_n)$$

where $\displaystyle a_n = \frac{1}{k\sin(1/k)}-1 >0.$ We know that $\displaystyle \prod_{k=1}^n v_n$ converges if and only if $\displaystyle \sum_{k=1}^n a_n$ converges. Using Taylor series for the $\sin$ term and expanding as geometric series we have $$a_n= \frac{1}{1-1/(6k^2) +\mathcal{O}(k^{-3})}-1= \frac{1}{6k^2} + \mathcal{O}\left(\frac{1}{k^3}\right)$$

so we find that $\sum a_n$ converges, so $\sum u_n$ diverges.

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  • $\begingroup$ Thanks a lot! I can't believe my instinctive reaction wasn't to use the ratio test! $\endgroup$ – Chon Jun 10 '12 at 11:41
  • $\begingroup$ Neither can I. The credit is all to N.S. for that idea. $\endgroup$ – Ragib Zaman Jun 10 '12 at 11:44
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$$\frac{u_{n+1}}{u_n}=\frac{(n+1)!\prod_{k=1}^{n+1} \sin\left(\frac{x}{k}\right)}{n!\prod_{k=1}^n \sin\left(\frac{x}{k}\right)}=(n+1)\sin(\frac{x}{n+1})$$

Thus, $\left|\lim_n \frac{u_{n+1}}{u_n}\right|=|x|$. Now, ratio test solves all cases excepting $x=1$.

If $x=1$, I think the product is known, but can't remember a reference.

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I tried to solve the case $x=1$

$$ \frac{u_{n+1}}{u_{n}}=(n+1)\sin \left(\frac{1}{n+1} \right)=(n+1) \left( \frac{1}{n+1}-\frac{1}{6(n+1)^3}+o(1/n^3) \right)=1+o(1/n)$$

Now: $$ v_{n}=\frac{1}{\sqrt{n}} $$

$$\frac{v_{n+1}}{v_{n}}=1-\frac{1}{2n}+o(1/n)$$

$$ \frac{u_{n+1}}{u_{n}}-\frac{v_{n+1}}{v_{n}}=\frac{1}{2n}+o(1/n)$$ There exists $ N$ such that $$ \forall n\geq N, \frac{u_{n+1}}{u_{n}} \geq \frac{v_{n+1}}{v_{n}}$$

$\sum v_{n}$ is divergent so $\sum u_{n}$ is divergent.

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For the case $x=1$, we can also from: $$ \frac{u_{n+1}}{u_{n}}=(n+1)\sin \left(\frac{1}{n+1} \right)=(n+1) \left( \frac{1}{n+1}-\frac{1}{6(n+1)^3}+o(1/n^3) \right)$$given by Chon , say that:$$ \frac{u_{n+1}}{u_{n}}= 1- \frac{1/6}{n} + o\left(\frac 1{n^2}\right)$$ and apply Raabe-Duhamel test, who say that the série diveges since $\frac 16 < 1$

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