# Determine convergence of the sequence $x_0=1 , x_{n+1}=x_n (1+ 2^{-(n+1)})$

I want to check if the following sequence converges:

$$x_0=1 , x_{n+1}=x_n \left(1+ \frac{1}{2^{n+1}}\right)$$

I proved the sequence is increasing :

$$\cfrac{x_{n+1}}{x_n}=1+ \cfrac{1}{2^{n+1}} \gt 1$$

Now I should prove it is bounded above. let's write some terms of the equation:

\begin{align} x_0&=1 \\[2ex] x_1&=1\cdot\left(1+\dfrac{1}{2^1}\right)\\[2ex] x_2&=\left(1+\dfrac{1}{2^1}\right)\cdot\left(1+ \dfrac{1}{2^2}\right)\\[2ex] x_3&=\left(1+\dfrac{1}{2^1}\right)\cdot\left(1+ \dfrac{1}{2^2}\right)\cdot\left(1+ \dfrac{1}{2^3}\right)\\[2ex] \end{align} So, we can write:

$$x_{n+1}=\left(1+\frac{1}{2^1}\right)\cdot\left(1+ \frac{1}{2^2}\right)\cdots\left(1+\frac{1}{2^{n}}\right)\cdot\left(1+\frac{1}{2^{n+1}}\right)$$

Here, I'm not sure how to prove it is bounded above.

• Is it $x_{n+1}=x_n\left(1+\dfrac1{2^{n+1}}\right)$ or $x_{n+1}=x_n\left(1+\dfrac1{2^n+1}\right)$? Sep 22 '20 at 14:59
• It is the first one. thank you I fixed it. Sep 22 '20 at 15:01
• Please rollback the edit I made if you found anything in the question that convinces you that the question has changed from the original in terms of content. Sep 22 '20 at 15:54
• @sai-kartik it is ok, thanks for the edit Sep 22 '20 at 16:00

First show that $$x_n \leq \sqrt{2}^{n+1}$$ through mathematical induction. Then, $$0 \leq x_{n+1} - x_n = x_n \frac{1}{2^{n+1}} \leq \frac{1}{\sqrt{2}^{n+1}}$$. You can easily check that given sequence is Cauchy sequence.

• It is really hard that the idea comes to my mind. How you thought about proving $x_n \leq \sqrt{2}^{n+1}$ then it helps to compare with Cauchy sequence? Sep 22 '20 at 16:53
• Idea : rate of increase of given sequence is very small, and $\sqrt{2}^{n+1}$ is quickly increasing(in terms of exponential function)
– Han
Sep 22 '20 at 16:56
• Observe that $x_0 \leq \sqrt{2}, x_1 \leq \sqrt{2}^2, x_2 \leq \sqrt{2}^3$. And Suppose $x_n \leq \sqrt{2}^{n+1}$ for $n>1$. Then $x_{n+1} = x_n + x_n / (2^{n+1})\leq \sqrt{2}^{n+1} + \frac{1}{\sqrt{2}^{n+1}} \leq \sqrt{2}^{n+2}$
– Han
Sep 22 '20 at 17:02
• Use $\sqrt{2} = 1.4\cdots$ in last inequality.
– Han
Sep 22 '20 at 17:06
• There is no specific reason for proving $x_n \leq \sqrt{2}^{n+1}$, but I focused on expressing the difference($x_{n+1} - x_n = \frac{x_n}{2^{n+1}}$) between two consecutive sequences in a given equation as a geometric sequence.
– Han
Sep 22 '20 at 17:16

You have $$x_n = \prod_{k=1}^n \left(1+\frac{1}{2^k}\right) = e^{\sum_{k=1}^n\ln \left(1+\frac{1}{2^k}\right)} \leq e^{\sum_{k=1}^n\frac{1}{2^k}} \leq e^{\sum_{k=1}^\infty\frac{1}{2^k}} = e$$ where for the first inequality we used that $$\ln(1+x)\leq x$$ for all $$x>-1$$. This shows the sequence is bounded.

The Limit Exists

Cross-multiplication and comparison shows that $$\frac{1+\frac1{2^{k-1}}}{1+\frac1{2^k}}\le1+\frac1{2^k}\le\frac{1-\frac1{2^k}}{1-\frac1{2^{k-1}}}\tag1$$ Then telescoping products lead to $$\underbrace{\prod_{k=1}^\infty\frac{1+\frac1{2^{k-1}}}{1+\frac1{2^k}}}_2 \le\prod_{k=1}^\infty\left(1+\frac1{2^k}\right) \le\underbrace{\overset{\substack{k=1\\\downarrow\\[6pt]\,}}{\frac32}\overbrace{\prod_{k=2}^\infty\frac{1-\frac1{2^k}}{1-\frac1{2^{k-1}}}}^2}_3\tag2$$ Therefore, $$x_n$$ is an increasing sequence that is bounded above by $$3$$, so $$\lim_{n\to\infty}x_n=\prod_{k=1}^\infty\left(1+\frac1{2^k}\right)\tag3$$ exists.

Bounding The Limit

Using $$(1)$$ and we get $$\left(1+\frac1{2^n}\right)\prod_{k=1}^n\left(1+\frac1{2^k}\right) \le\prod_{k=1}^\infty\left(1+\frac1{2^k}\right) \le\frac1{1-\frac1{2^n}}\prod_{k=1}^n\left(1+\frac1{2^k}\right)\tag4$$ The greater the $$n$$ used, the tighter the bounds in $$(4)$$.

The Limit

Using $$(4)$$ with $$n=30$$, we get that $$2.38423102903137172\color{#C00}{35}\le\prod_{k=1}^\infty\left(1+\frac1{2^k}\right)\le2.38423102903137172\color{#090}{55}\tag5$$