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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,

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##?

$$

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##?