# convergence of $(1+\frac{1}{2})(1+\frac{1}{4})\ldots(1+\frac{1}{2^n})$

I was given a problem of testing sequence convergence. The sequence is defined as:

$$x_n= (1+\frac{1}{2})(1+\frac{1}{4})\ldots(1+\frac{1}{2^n})$$ My first idea was to define $y_n$ as follows: $$y_n = \log{x_n} = \log{(1+\frac{1}{2})}+\log{(1+\frac{1}{4})}+\ldots+\log{(1+\frac{1}{2^n})}$$ Then, out of few methods i know for proving convergence, i decided to use the Cauchy's criteria i.e. for some sufficiently large natural numbers $n$ and $p$ such that $p \gt n$, we have (from triangle inequality $|y+x| \le |y|+|x|$):

$$| y_{n+p} - y_n |= | \log{(1+\frac{1}{2^{n+1}})} |+ | \log{(1+\frac{1}{2^{n+2}})} |+\ldots+ |\log{(1+\frac{1}{2^{n+p}})}|$$ Since the argument logarithm in every term has to be greater than $1$, we can ignore the absolute values on the right side and write the following:

$$| y_{n+p} - y_n |= \log{\bigg{(} (1+\frac{1}{2^{n+1}})(1+\frac{1}{2^{n+2}}) \ldots (1+\frac{1}{2^{n+p}})} \bigg{)}$$

This is where i get stuck, since the idea here is to prove the term in brackets is less than $e$, but I fear that's as difficult as solving the task. Is this a good approach? If not, what would be a better one?

• See here. – sciona Jan 25 '15 at 10:00
• Hint: $\ln(1+x)\leqslant x$ for $x>0.$ – M. Strochyk Jan 25 '15 at 10:02

## 3 Answers

We will prove a more general statement:

If $a_n$ is a positive sequence then $\prod (1+a_n)$ converges iff $\sum a_n$ converges.

Proof: We have $$\log\left(\prod (1+a_i)\right) = \sum \log(1+a_i) \leq \sum a_i$$

where we have used $\log(1+x)\leq x$ which is valid for all $x > -1$. The reversed implication is proven here.

Now take $a_n = \frac{1}{2^n}$. Since $\sum \frac{1}{2^n} = 1$ it follows that $\prod\left(1+\frac{1}{2^n}\right) = \left(1 + \frac{1}{2}\right)\left(1 + \frac{1}{4}\right)\left(1 + \frac{1}{8}\right)\cdots$ converges.

If you really want to use Cauchy's criterion then by using $\log(1+x)\leq x$ we get

$$|y_{n+p} - y_n| \leq \sum_{k=1}^p \frac{1}{2^{n+k}} = \frac{1}{2^n}(1-2^{-(p+1)}) < \frac{1}{2^{n}}$$

Given $\epsilon>0$ then if $n>\log_2(1/\epsilon)$ we get $|y_{n+p} - y_n| < \epsilon$ for all $p>0$.

• The statement is actually an if and only if statement. – IAmNoOne Jan 25 '15 at 10:04
• @Nameless Very true, I removed the 'f' to avoid adding having to do the other implication:) But for completeness I will add it. – Winther Jan 25 '15 at 10:05

Fine. So you already have the series

$$\sum_{n=1}^\infty\log\left(1+\frac1{2^n}\right)$$

Define for $\;x\ge0\;$:

$$f(x)=\log(1+x)- x\implies f'(x)=\frac1{1+x}-1=\frac{-x}{x+1}<0\implies\;f(x)$$

is monotone descending, and thus

$$\forall x\ge 0\;,\;\;f(x)\le f(0)=0$$

and we're done.

Finally: use the comparison test for infinite positive series

Here's an approach that doesn't rely on logarithms. If $a_1,\ldots,a_n$ are numbers in $[0,1)$ with sum less than $1$ then the following inequality holds: $$\prod_{k=1}^n(1+a_k)\leq\frac{1}{1-\sum_{k=1}^na_k}.$$ This follows from $$1+a\leq\frac{1}{1-a}$$ for $a\in[0,1)$ and induction. In your case this shows that $$\prod_{k=1}^n(1+2^{-k})=\frac{3}{2}\prod_{k=2}^n(1+2^{-k})\leq 3.$$ Now a bounded increasing sequence converges.