Problem: Show that, for $n\in \mathbb{N}$, the sequence


is convergent.

The sequence should converge if the general factor (last factor) approaches 1. So we can form the function

$$f(x) = 1-\frac{1}{2^x},$$

only by looking at this, it's abundantly clear that $1/2^x\rightarrow 0$ as $x\rightarrow \infty.$ This means that


and convergence is shown.

This seems a bit too easy to be enough of a proof. Does this suffice or am I missing something critical?

  • $\begingroup$ What you are actually doing is showing that the infinite product $$\prod_{n=1}^{+ \infty} \left(1-\frac{1}{2^n}\right)$$ converges. Just like infinite series, you have to be careful what you are doing. Saying that the last term converges to zero is NOT sufficient to say the whole thing converges to zero. $\endgroup$ – user370967 Dec 15 '17 at 9:17
  • $\begingroup$ @Math_QED: Apologies, I meant last factor approaches to 1. But if the general term approaches 1, shouldn't the entire thing be convergent? Can you please show me a more stringent proof? $\endgroup$ – Parseval Dec 15 '17 at 9:20
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    $\begingroup$ Just for curiosity's sake: turns out the sequence converges to $\approx0.289$. $\endgroup$ – Peter Woolfitt Dec 15 '17 at 9:24
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    $\begingroup$ More curiosity: This is a special instance of the Euler_function or Q-Pochhammer symbol, the limit is $0.288788\dots$ $\endgroup$ – gammatester Dec 15 '17 at 9:31

I would say that it is that easy, but your approach isn't the right one. Rather, this is a monotonously decreasing sequence bounded by $0$, so it must converge. Whether that limit is $0$ or something else requires some more careful calculation, though.

To see why your approach is flawed in general, consider what would happen if there were a $+$ inside the brackets instead of a $-$. In that case, you would have a product of terms that come closer and closer to $1$, but with more and more terms. There is no easy way to conclude whether that converges or diverges. For instance, $$b_n=\left(1+\frac11 \right)\left(1+\frac12 \right)\cdots\left(1+\frac1n \right)=n+1$$ diverges.

  • $\begingroup$ Thanks Arthur, I understand why my approach doesn't hold. What whould be the start of a correct approach? $\endgroup$ – Parseval Dec 15 '17 at 9:22
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    $\begingroup$ @Parseval I told you in my first paragraph: the sequence is decreasing, but all terms are positive (you have to show these two things, of course). That necessarily means that it converges to something. $\endgroup$ – Arthur Dec 15 '17 at 9:24
  • $\begingroup$ Yes, but this is a 7 point question on an old exam. I highly doubt that the sentence "this is a monotonously decreasing sequence bounded by 0, so it must converge." warrants all 7 points. I have to back up that statement. Maybe use the derivative? $\endgroup$ – Parseval Dec 15 '17 at 9:25
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    $\begingroup$ @Parseval As I said, yes, you do have to show that it is decreasing. For instance, $a_{n+1} = a_n \cdot \left(1-\frac1{2^{n-1}}\right)$. Can you see how $a_{n-1}<a_n$ for all $n$ follows from there? You also have to show that it is bounded below. Can you see how to prove (by induction, for instance) that $a_n>0$? Derivatives are no use, since $n$ is not a continuous variable. How would you do $\lim_{h\to 0}\dfrac{a_{n+h}-a_n}{h}$? What does $a_{n+h}$ mean when $h$ is a very number (i.e. not an integer) $\endgroup$ – Arthur Dec 15 '17 at 9:36
  • $\begingroup$ Thank you, I will ponder your questions a bit and return with some followup questions. $\endgroup$ – Parseval Dec 15 '17 at 9:41

In order to determine if $(a_n)_{n\in\mathbb N}$ converges to $0$ or not, you can check if $\lim_{n\to\infty}\log a_n=-\infty$ or not. Note that$$\log a_n=\log\left(1-\frac12\right)+\log\left(1-\frac1{2^2}\right)+\cdots+\log\left(1-\frac1{2^n}\right).$$So, we are dealing with a series here. The terms are negative, but it converges to $-\infty$ if and only if the series$$\sum_{n=1}^\infty-\log\left(1-\frac1{2^n}\right)\tag{1}$$converges to $+\infty$. But$$\lim_{x\to0}\frac{-\log(1-x)}x=1$$and therefore the divergence of the series $(1)$ is equivalent to the divergence of the series $\sum_{n=1}^\infty\frac1{2^n}$. But this series converges. Therefore, $\lim_{n\to\infty}a_n>0$.

  • $\begingroup$ I don't understand how you're using that last limit equal to 1. Also, is there any other way of doing this using differentiation? We have yet not covered series in this course. $\endgroup$ – Parseval Dec 15 '17 at 9:35
  • $\begingroup$ @Parseval The fact that the limit is $1$ is irrelevant. What matters here is that the limit is a real number greater than $0$. This is just the comparison test. It is easy to check that $\frac{-\log(1-x)}x$ is bound by some constant $K$ in $\left[\frac12,1\right)$. Therefore,$$\log a_n\geqslant-K\left(\frac12+\frac1{2^2}+\cdots+\frac1{2^n}\right)$$for each $n\in\mathbb N$. $\endgroup$ – José Carlos Santos Dec 15 '17 at 9:45

Correct me if wrong.

$a_n \gt 0$, $ n \in \mathbb{Z+}.$

$a_{n+1} = a_n (1-\dfrac{1}{2^{n+1}}) \lt a_n.$

Hence $a_n$, positve, is strictly monotonically decreasing , bounded below.

$\rightarrow :$


  • $\begingroup$ Could you please elaborate on how your conclusions follows from $a_{n+1}=a_n(1-1/2^{n+1})<a_n$? $\endgroup$ – Parseval Dec 15 '17 at 9:52
  • $\begingroup$ Parseval: 1-1/2^{n+1} < 1, correct? Hence by replacing the parenthesis by 1 we make it bigger and get 1×a_n .Does this make sense? $\endgroup$ – Peter Szilas Dec 15 '17 at 9:58
  • $\begingroup$ Yes it does. But why would you want to make it bigger? Say we replace it by one, then we get $a_{n+1}=a_n\cdot.$ but this is not really true. $\endgroup$ – Parseval Dec 15 '17 at 10:03
  • $\begingroup$ We want to show : a_{n+1} smaller!! than a_n. I hope this inequality shows it. If true : bounded below (by 0),mon. decreasing,i.e. a_{n+1} less than a_{n}, it follows: convergent.(a theorem , in textbooks under monotonically sequence). Let me know if you agree . $\endgroup$ – Peter Szilas Dec 15 '17 at 10:22
  • $\begingroup$ Example: a_1=1/2; a_2 = a_1(1-1/2^2) = 3/4 a_1. Does it follow that a_2 less than a_1? $\endgroup$ – Peter Szilas Dec 15 '17 at 10:29

First, the sequence $a_n$ is monotone $$ a_1>a_2>a_3>\cdots >a_n>\cdots $$ Second, the sequence $a_n$ is limited. For $0<x<1$ we have $ln(1-x)<x$. Then $$ \sum_{k=1}^{n}\ln\left(1-\frac{1}{2^k}\right) \leq \sum_{k=1}^{n}\frac{1}{2^k} $$ and $$ |a_n|=|e^{\log a_n}| = \left|\;e^{\sum_{k=1}^{n}\ln\left(1-\frac{1}{2^k}\right)}\;\right| \leq \left|e^{\sum_{k=1}^{n}\frac{1}{2^k}}\right| = e^{\frac{1/2-(1/2)^n}{1-1/2}} \leq e^{\frac{1/2}{1-1/2}}=e $$


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