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The question is this.

Let $(s_n)$ be a sequence such that $$\left|s_{n+1}-s_n\right| < 2^{-n}, \forall n \in\mathbb N$$ Prove that $(s_n)$ is a Cauchy sequence and hence convergent.

My proof is below.

Proof. Let $\epsilon > 0 $ and $N = \dfrac{\ln\left(\frac{1}{\epsilon}\right)}{\ln(2)}$. Then, $\forall n > N = \dfrac{\ln(\frac{1}{\epsilon})}{\ln(2)}$ implies $$n\ln(2) > \ln\left(\frac{1}{\epsilon}\right)$$ $$2^n > \frac{1}{\epsilon}$$ $$\epsilon > \frac{1}{2^n} > \left|s_{n+1}-s_n\right|$$Q.E.D.

Is this ok??


marked as duplicate by Jendrik Stelzner, Thomas Shelby, José Carlos Santos real-analysis May 30 at 13:18

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    $\begingroup$ No, you have not shown that the sequence is Cauchy. Write down the definition and use the triangle inequality along with the appropriate telescoping sequence and the given bound to show that the sequence is Cauchy. $\endgroup$ – copper.hat Sep 11 '13 at 2:08
  • $\begingroup$ This is correct proof of the fact that $s_n$ is convergent, but it's not the proof of the fact that $s_n$ is a Cauchy sequence. While these two are equivalent, in this problem they ask for a specific proof. As for how to solve it, use the fact that $2^{-n} + 2^{-n-1} + \ldots + 2^{-n-k} = 2^{-n}(2^0 + 2^{-1}+ ... + 2^{-k}) = 2^{-n}\cdot \mathrm{geometric series}$ $\endgroup$ – xyzzyz Sep 11 '13 at 2:09
  • $\begingroup$ The definition in my text book is this!! $\forall \epsilon > 0, \exists N \ s.t\ m,n > N \Rightarrow |s_n - s_m| < \epsilon$. Can't I just replace $n+1$ by $n$ and $n$ by $m$?? $\endgroup$ – eChung00 Sep 11 '13 at 2:12
  • $\begingroup$ no. You cannot arbitrary replace $n+1$ and $n$ by $n$ and $m$, since the first two have the property that they are $1$ unit apart, whereas the theorem say that $n$ and $m$ can by anything bigger than $N$, so $m$ can be very far off from $n$, and you should still have that they are closer than $\epsilon$. $\endgroup$ – Daniel Montealegre Sep 11 '13 at 2:16
  • $\begingroup$ What happens if we have $|s_{n+1}-s_n|< \frac{1}{n}$? $\endgroup$ – Mario De León Urbina Apr 14 '14 at 19:07

No. It does not work. You have to show that given $\epsilon>0$ then you can find $N$ big enough so that if $n,m\geq N$ then $|s_n-s_m|<\epsilon$. You showed that $s_n$ and $s_{n+1}$ have a difference of less than $\epsilon$ but you want to show that $s_n$ and $s_m$ have a difference of less than $\epsilon$, see the difference?

Try this. Fix $\epsilon>0$. Choose $N$ big enough an integer so that $1/2^N<\epsilon$ (this is possible, why?). Then if $n,m> N$ (say wlog that $m\geq n$, then $$|s_n-s_m|=|s_n-s_{n+1}+s_{n+1}-...+s_{m+1}-s_{m}|$$Apply the triangle inequality, and try to go from here. Let me know if you get stuck.


So you can finish it like this: $$ |s_n-s_m|\leq |s_n-s_{n+1}|+...+|s_{m-1}-s_m|\leq 1/2^n+...+1/2^{m-1}$$$$=(1/2^n)(1+1/2+...+1/2^{m-n-1})<(1/2^n)(2)=1/2^{n-1} $$ Since $n>N$ then $n-1\geq N$ so $2^{n-1}\geq 2^N$ which implies that $1/2^{n-1}\leq 1/2^N<\epsilon$ just as we wanted.

  • $\begingroup$ Choose N so that the sum above N is less than epsilon actually $\endgroup$ – Deven Ware Sep 11 '13 at 3:39
  • $\begingroup$ Thanks for your comment!! I thought about what you said and tried this!!$|s_n-s_m| = | s_n-s_{n+1}+s_{n+1}-s_m| \leq |s_n-s_{n+1}| + |s_{n+1}-s_m| < |s_n-s_{n+1}| < 2^{-n}$. Is this ok?? I am so confused with this epsilon proof!!lol $\endgroup$ – eChung00 Sep 11 '13 at 17:38
  • $\begingroup$ @DevenWare I meant to write $n,m> N$ not "$n,m\geq N$". It should work now. $\endgroup$ – Daniel Montealegre Sep 11 '13 at 22:23
  • $\begingroup$ It worked the whole time, I just felt like it would be clearer to say the way I did $\endgroup$ – Deven Ware Sep 12 '13 at 2:22
  • $\begingroup$ @DevenWare Reported $\endgroup$ – Daniel Montealegre Sep 12 '13 at 5:02

The following result is useful

if $\sum_{n=1}^{\infty}|a_{n+1}-a_n|< \infty$, then ${a _n} $ is a Cauchy sequence.

For a proof of this result, see here.


\begin{align} &\mbox{Let's}\quad m >n \\ \left\vert s_{m} - s_{n}\right\vert &= \left\vert \left(s_{m} - s_{m - 1}\right) + \left(s_{m - 1} - s_{m - 2}\right) + \cdots + \left(s_{n + 1} - s_{n}\right) \right\vert \leq \sum_{i = n}^{m - 1}\left\vert s_{i + 1} - s_{i}\right\vert < \sum_{i = n}^{m - 1}2^{-i} \\[3mm]&= {2^{-n}\left\lbrack 2^{-\left(m - n\right)} - 1\right\rbrack \over 1/2 - 1} = 2^{1 -n}\left\lbrack 1 - 2^{-\left(m - n\right)}\right\rbrack < 2^{1 - n} \end{align}

Given $\displaystyle{\epsilon > 0}$, just take any $\displaystyle{N > 1 - {\ln\left(\epsilon\right) \over \ln\left(2\right)}}$. Then, $\displaystyle{\quad m, n > N\quad\Longrightarrow\quad \left\vert s_{m} - s_{n}\right\vert < \epsilon}$.

  • $\begingroup$ I think you messed up in a power. I get $1/2^{n-1}$, not $1/2^{n+1}$. $\endgroup$ – Daniel Montealegre Sep 11 '13 at 22:25
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    $\begingroup$ Thanks @DanielMontealegre . I just checked it. Thewe was a minor error. $\endgroup$ – Felix Marin Sep 11 '13 at 22:32

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