lim-n-r-1-n-1-2-r-where-greatest-integer-finction-

Question Number 213511 by universe last updated on 07/Nov/24

\lim_{n \to \infty} [\underset{r = 1}{\sum^{n}} \frac{1}{2^{r}}]

where [∙] greatest integer finction

Answered by issac last updated on 07/Nov/24

\underset{h = 1}{\sum^{M}} 2^{- h} \leq 1

\lim_{M \to \infty} \underset{h = 1}{\sum^{M}} 2^{- h} \leq 1 but \underset{h = 1}{\sum^{\infty}} 2^{- h} = 1

cus 0.99999 \dots \dots = 1

⌊ \lim_{M \to \infty} \underset{h = 1}{\sum^{M}} 2^{- h} ⌋ = 1

Commented by mr W last updated on 07/Nov/24

t h e q u e s t i o n i s

\lim_{n \to \infty} ⌊ \times ⌋ = ?

n o t ⌊ \lim_{n \to \infty} \times ⌋ = ?

Answered by lepuissantcedricjunior last updated on 09/Nov/24

\lim_{n \to + \infty} [\underset{r = 1}{\sum^{n}} \frac{1}{2^{r}}] = \lim_{n \to + \infty} [- 1 + \underset{r = 0}{\sum^{n}} \frac{1}{2^{r}}]

= \lim_{n \to + \infty} [- 1 + \frac{1 - (\frac{1}{2^{n + 1}})}{1 - \frac{1}{2}}]

= \lim_{n \to + \infty} [- 1 + 2 (1 - 2^{- (n + 1)})]

= [- 1 + 2] = 1

Commented by mr W last updated on 09/Nov/24

w r o n g!

Answered by mr W last updated on 07/Nov/24

\underset{r = 1}{\sum^{n}} \frac{1}{2^{r}} = \frac{\frac{1}{2} (1 - \frac{1}{2^{n}})}{1 - \frac{1}{2}} = 1 - \frac{1}{2^{n}} < 1, b u t > 0

\Rightarrow [\underset{r = 1}{\sum^{n}} \frac{1}{2^{r}}] = 0

\Rightarrow \lim_{n \to \infty} [\underset{r = 1}{\sum^{n}} \frac{1}{2^{r}}] = \underset{n \to \infty}{\lim 0} = 0 ✓

Commented by issac last updated on 07/Nov/24

D a m n i w a s w r o n g

Commented by universe last updated on 07/Nov/24

t h a n k s s i r

Answered by Berbere last updated on 07/Nov/24

[x] < x

\Rightarrow 0 ⩽ [\underset{r = 1}{\sum^{n}} \frac{1}{2^{r}}] ⩽ \underset{r = 1}{\sum^{n}} \frac{1}{2^{r}} = 1 - {(\frac{1}{2})}^{n} < 1

\Rightarrow \forall n \in N [Σ \frac{1}{2^{r}}] = 0

Commented by universe last updated on 07/Nov/24

t h a n k y o u s o m u c h s i r