# Find infimum, supremum and max, min of a set if they exist

I need to find infimum, supremum and max, min of a set if they exist:$$E=\left\{1-\sum_{m=1}^{n}\frac{1}{4^m}\ : n\in \mathbb{N}\right\}$$ It seems that $$\max E=\sup E=\frac{3}{4}$$ and $$\inf E=\frac{2}{3}$$.

However, I am not sure if I can claim that $$\min E=\frac{2}{3}$$.

Intuitively, I'm certain that this set does not have a minimum value.

• Confirm... do you mean to have $\sum\limits_{\color{red}{m}=1}^{\color{blue}{n}}\dfrac{1}{4^\color{blue}{n}}$? Or do you mean to have $\sum\limits_{\color{red}{m}=1}^{\color{blue}{n}}\dfrac{1}{4^\color{red}{m}}$? Oct 28, 2020 at 15:32
• I changed the index. Thank you for noticing. Oct 28, 2020 at 15:36

Consider the sequence $$a_n=1-\sum_{m=1}^n\frac{1}{4^m}$$ Note that $$E=\{a_n\}_n$$, where that sequence is strictly decreasing and bounded. Therefore you have a maximum (because its strictly decreasing) and an infimum with no minimum because if it ever reached the minimum (suppose it's reached by $$a_{n_0}$$), it contradicts the fact it is the minimum because the sequence is strictly decreasing.

Okay.

So if $$S_n = 1-\sum_{m=1}^n \frac 1{4^n}$$ its easy to see that $$S_{k+1} = S_k -\frac 1{4^{k-1}} < S_k$$ so $$S_1 >S_2 > S_3.....$$.

So every $$S_k \le S_1$$ and so $$\sup S_n = \max S_n = S_1$$.

And because, for every $$S_k$$ there is always $$S_{k+1} < S_k$$ we can't have have a smallest $$S_k$$ so $$\min S_n$$ can not exists.

So what remains to answer are: Are the $$S_k$$ bounded below and if so what is $$\inf S_n$$?

Are you familiar with geometric series?

$$1 + r + r^2 + r^3 +........ + r^n = M_n$$

Then $$M_n(1-r) = (1 + r + r^2 + r^3 +........ + r^n) - (r+r^2 + ...... + r^n + r^{n+1}) = 1-r^{n+1}$$ so

$$M_n = \frac {1-r^{n+1}}{1-r}$$.

So $$\sum_{m=1}^n \frac 14 = (\sum_{m=0}^n \frac 14) -1=\frac {1-(\frac 14)^{n+1}}{1-\frac 14} -1=$$

$$\frac {4(1-(\frac 14)^{n+1}}3 -1= \frac {4-(\frac 14)^n}3 - 1=\frac {1-\frac 1{4^n}}3$$

So $$S_n = 1-\frac {1-\frac 1{4^n}}3= \frac {2+\frac 1{4^n}}3$$.

And as $$\frac 1{4^n} >0$$ we can see than indeed $$S_n > \frac 23$$ and the sequence is bounded below by $$\frac 23$$.

If we take any number larger than $$\frac 23$$, say $$\frac 23 + \iota$$ then we can find a $$\frac 1{4^n}$$ so that $$\frac 23 < \frac {2+\frac 1{4^n}}3 < \frac 23 + \iota$$. (Why?)

So any number larger than $$\frac 23$$ is not a lower bound, so $$\frac 23$$ is the greatest lower bound.

And $$\inf S_n = \frac 23$$.