If ##|s_{n+1} - s_n| \lt 1/2^n##, then ##(s_n)## is a Cauchy sequence

[Hall]

Homework Statement

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Relevant Equations

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My attempt: It can be proved that $\lim \frac{1}{2^n} = 0$. Consider, $\frac{\varepsilon}{k} \gt 0$, there exists $N$, such that
$$
n \gt N \implies \frac{1}{2^n} \lt \varepsilon
$$
Take any $m,n \gt N$, and such that $m - k = n$.
$|s_m - s_{m-1} | \lt \frac{1}{2^{m-1}} \lt \frac{\varepsilon}{k}$
$| s_{m-1} - s_{m-2} | \lt \frac{1}{2^{m-2}} \lt \frac{\varepsilon}{k}$
$| s_{m-2} - s_{m-3} \ \lt \frac{1}{2^{m-3} } \lt \frac{\varepsilon}{k}$
$\vdots$
$| s_{m-k+1} - s_{m-k} | \lt \frac{1}{2^{m-k} } \lt\frac{\varepsilon}{k}$
By using triangle inequality repeatedly, we have
$| s_m - s_{m-k} | \lt \varepsilon$
$| s_m - s_n| \lt \varepsilon$

But my doubt with this method is that my choice of $\frac{\varepsilon}{k}$ was not, I mean, very arbitrary, that $k$ determined $m$ and $n$.

And why everyone else on internet is doing it by using the triangle inequality and making it less than $\sum_{k=n}^{m-1} \frac{1}{2^k}$ and not to $\varepsilon$?

 

[PeroK]

What about $m = n +k +1$?

 

[WWGD]

The larger you make/consider your tail, the larger the differences, which are then not neccesarily staying within $\epsilon$ from each other. Maybe using $|S_{n+1} -S_{n}| < \epsilon/2^{n}$ could work.

 

[Hall]

The larger you make/consider your tail, the larger the differences, which are then not neccesarily staying within $\epsilon$ from each other. Maybe using $|S_{n+1} -S_{n}| < \epsilon/2^{n}$ could work.

Can you please explain a little more?

 

[Office_Shredder]

The worst case for you is if the sequence is strictly increasing, and each $s_{n+k}$ tries to get as far away from $s_n$ as possible. How far away can it go?

 

[Hall]

The worst case for you is if the sequence is strictly increasing, and each $s_{n+k}$ tries to get as far away from $s_n$ as possible. How far away can it go?

$\sum_{j=n}^{n+k-1} \frac{1}{2^j}$

 

[Office_Shredder]

Can you write down an upper bound for that (let k go to infinity)?
Then use that to say something about the sequence being cauchy.

 

[Hall]

Can you write down an upper bound for that (let k go to infinity)?
Then use that to say something about the sequence being cauchy.

The maximum distance between any two elements of sequence $(s_n)$, can be found by letting $k \to \infty$ in the sum above (which is the same thing to say that the distance between $s_n$ and a very far off term), so, we have (the notation $s_{\infty}$ might not be very okay)
$$
| s_{\infty} - s_n| \lt \sum_{n}^{\infty} \frac{1}{2^j}$$
$$
| s_{\infty} - s_n| \lt \frac{1}{2^{n-1} }$$

$$
\because \lim \frac{1}{2^n} = 0$$
$$
\therefore n \gt N \implies \frac{1}{2^n} \lt \varepsilon$$

Take any $m,n \gt N+1$, we have established above that no matter how far away $m$ is from $n$, we have an upper bound, so,
$$
| s_m - s_n| \lt \frac{1}{2^{n-1} }$$
As $n \gt N+1$, the least value $n$ can take is $N+2$, therefore RHS of the above inequality would become $\frac{1}{2^{N+1} }$ which is, of course, is less that $\epsilon$.
$$
| s_m - s_n| \lt \varepsilon $$

 

[Office_Shredder]

I think you have the right idea, but it's not well written. The first line should probably look like: given $\epsilon$, pick $N$ such that $\frac{1}{2^N} < \epsilon$

 

[Hall]

I think you have the right idea, but it's not well written. The first line should probably look like: given $\epsilon$, pick $N$ such that $\frac{1}{2^N} < \epsilon$

Let me try to re-write it.

For a given $\varepsilon \gt 0$, there exists $N$ such that $n \gt N \implies \frac{1}{2^n} \lt \varepsilon$.

Take any $m, n \gt N+1$, and we can safely assume $m-n =k$ for some positive $k$. So, we have

$|s_m - s_{m-1} | \lt \frac{1}{2^{m-1}}$
$| s_{m-1} -s_{m-2}| \lt \frac{1}{2^{m-2} }$
$\vdots$
$|s_{m-k+1} - s_n| \lt \frac{1}{2^n}$
By using triangle inequality repeatedly, we have
$| s_m - s_n| \lt \sum_{j= n}^{m-1} \frac{1}{2^j} \lt \sum_{n}^{\infty} \frac{1}{2^j}$
$|s_m - s_n| \lt \frac{1}{2^{n-1} } \lt \varepsilon$.

Hence, $(s_n)$ is a Cauchy sequence.