-3

Let $c\in \mathbb{R},c>0$ and define the sequence $(a_n)_{n\in \mathbb{N}}$ by $a_n:=\prod_{k=2}^n(1-\frac{1}{k^{1+c}})$. Does $\lim_{n\to\infty} a_n=0$ hold?

user26857
  • 52,094
bjn
  • 923
  • @GEdgar: I'm sorry, I want to exclude the trivial case with a factor equal to zero and the product runs through the values of $k=2$ to $n$ – bjn Apr 24 '15 at 19:59
  • Hands-on, purely real, approach: for every $x$ small enough, $$e^{-2x}\leqslant1-x\leqslant e^{-x}.$$ – Did Apr 24 '15 at 20:30

1 Answers1

0

The infinite product $$ \prod \left(1-\frac{1}{k^{1+c}}\right) $$ converges absolutely, which means the series $$ \sum\frac{1}{k^{1+c}} $$ converges absolutely.

So a theorem says that, if you ignore any factor which is equal to zero, then the partial products $$ \prod_{k=2}^n\left(1-\frac{1}{k^{1+c}}\right) $$ converge to a non-zero value.

For the theorem, consult a text on complex analysis. It seems watered-down presentations (like Wikipedia) do not include this...

GEdgar
  • 111,679
  • Thanks for your answer. I'm surprised that complex analytic methods are needed, because all terms are real. Has this theorem a specific name? – bjn Apr 24 '15 at 20:04
  • 1
    Complex analytic methods are not required. But it is included in complex analysis texts, since they cannot assume the reader already knows it, but it is used in complex analysis. – GEdgar Apr 24 '15 at 20:06
  • 1
    The proof is basically: for a complex number close enough to zero, we have $|z|/2 < |\log(1+z)| \le |z|$. References are listed here: http://math.stackexchange.com/questions/158089/infinite-products-reference-needed – GEdgar Apr 24 '15 at 20:09