Let $\mathbf{a_n = (1-\frac{1}{\sqrt2})...(1-\frac{1}{\sqrt{n+1}})}$, $n \ge 1$. Does $\lim_{n\to \infty} a_n$ exist? If so, what is it's value?
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Note that your sequence is bounded and monotonically decreasing – cQQkie Apr 22 '14 at 18:13
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1Hint: $1 - \frac{1}{\sqrt{n+1}} \le 1 - \frac{1}{n+1}$ – achille hui Apr 22 '14 at 18:15
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Hint: $\ln(1-x)<-x$ for $0<x<1$. – Hagen von Eitzen Apr 22 '14 at 18:19
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If $b_1, \ldots, b_n \in (0,1)$ then $$0<(1-b_1)(1-b_2)\dotsc(1-b_n) < \frac{1}{1+b_1+\ldots+b_n}.$$ This can be proved by induction. In this case $b_k=1/\sqrt{k+1}$. Try to find the limit of the right hand side.
WimC
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Generally speaking, $\displaystyle\prod(1\pm a_n)$ converges if and only if $\displaystyle\sum a_n$ converges. In this case, $\displaystyle\sum a_n$ is greater than the harmonic series, which is known to diverge.
Lucian
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Note that infinite products generally use a different definition of 'diverge' than series - in this case, the product does in fact have a limit (but that limit is such that we talk about it diverging). – Steven Stadnicki Apr 22 '14 at 22:34
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Note that if $k\ge 1$ then $\sqrt{k+1}<k+1$, then $\frac{1}{\sqrt{k+1}}>\frac{1}{k+1}$ and $1-\frac{1}{\sqrt{k+1}}<1-\frac{1}{k+1}=\frac{k}{k+1}$. Then \begin{align} a_n=\left(1-\frac{1}{\sqrt{2}}\right)\left(1-\frac{1}{\sqrt{3}}\right)\cdot\ldots\cdot\left(1-\frac{1}{\sqrt{n+1}}\right)&<\left(\frac{1}{2}\right)\left(\frac{2}{3}\right)\cdot\ldots\cdot\left(\frac{n}{n+1}\right)=\frac{1}{n+1} \end{align} Since $0<a_n<\frac{1}{n+1}$ for all $n$ we obtain $0\le \displaystyle{\lim_{x\rightarrow\infty}{a_n}}\le \displaystyle{\lim_{x\rightarrow\infty}}{\frac{1}{n+1}}=0$. Therefore $\displaystyle{\lim_{x\rightarrow\infty}{a_n}} =0$.
Ángel Mario Gallegos
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