# Limit of a converging sequence

Let the following sequence:$$u_0=1, \, \forall n\in\mathbb{N},\,u_{n+1}=\sqrt{u_n^2+\dfrac{1}{2^n}}$$

I try to find its limit. Well we can prove that $\forall n\in\mathbb{N},\,u_{n+1}-u_n\le \dfrac{1}{2^n}$ and so:$$\forall n\in\mathbb{N},\, u_n\le u_0+\sum_{k=0}^{n-1}\dfrac{1}{2^k}=1+2\left(1-\dfrac{1}{2^n}\right)\le3$$

So $(u_n)$ is bounded from above. Since it's an increasing sequence (easy to prove) then $(u_n)$ is converging.

I first thought that the limit is 3. To be sure I used Matlab and calculated $u_{1000}$ and $u_{2000}$. The result was $1.4142...$ so the limit is $\sqrt{2}$. I tried to prove it using the squeeze theorem (the only think I can think about) but I failed. Could you please help me?

• If $u_0=1$, then $u_1=\sqrt{1^2+\frac1{2^0}}=\sqrt2$, and the limit can’t be $\sqrt2$. Nov 10, 2014 at 11:56
• @Brian M.Scott :o so I think I did an error in Matlab. Nov 10, 2014 at 12:18

Concentrate on $u_n^2$ instead of on $u_n$, and unwind the recurrence:

\begin{align*} u_n^2&=u_{n-1}^2+\frac1{2^{n-1}}\\ &=u_{n-2}^2+\frac1{2^{n-2}}+\frac1{2^{n-1}}\\ &\;\vdots\\ &=u_{n-k}^2+\sum_{i=1}^k\frac1{2^{n-i}}\;. \end{align*}

Now carry it to $k=n$ to get a closed form for $u_n^2$, and take the limit as $n\to\infty$.

• Thank you! We get it like this. Nov 10, 2014 at 12:27
• @Scientifica: You’re welcome! Nov 10, 2014 at 12:39

From recursive definition of $a_n$, we get $$a_{n+1}^2=a_n^2+\frac{1}{2^n}$$ so $$a_n^2=a_0^2+1+\frac{1}{2}+\frac{1}{2^2}+\cdots+\frac{1}{2^{n-1}}=a_0^2+2-\frac{1}{2^{n-1}}.$$

Therefore $$a_n=\sqrt{3-1/2^{n-1}}.$$

• Thank you for your answer. Then the limit is $\sqrt{3}$. Nov 10, 2014 at 12:26

Think about this: $u_{n+1}^2=u_n^2+\frac12^n$

• Thank you! We get it like this. Nov 10, 2014 at 12:27