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.

  • 1
    $\begingroup$ 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 $\sum\limits_{\color{red}{m}=1}^{\color{blue}{n}}\dfrac{1}{4^\color{red}{m}}$? $\endgroup$
    – JMoravitz
    Oct 28, 2020 at 15:32
  • $\begingroup$ I changed the index. Thank you for noticing. $\endgroup$
    – edionz
    Oct 28, 2020 at 15:36

2 Answers 2


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.



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 $\inf S_n = \frac 23$.


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