Find-the-limits-when-n-goes-to-infinty-of-the-following-summation-series-a-1-n-2-k-1-n-E-kx-x-R-b-k-0-n-n-k-1-

Question Number 99685 by Ar Brandon last updated on 22/Jun/20

Find the limits when n goes to infinty of the following

summation series;

a ∖ \frac{1}{n^{2}} \underset{k = 1}{\sum^{n}} E (kx), x \in R

b ∖ \underset{k = 0}{\sum^{n}} {(\begin{matrix} n \\ k \end{matrix})}^{- 1}

Commented by Rasheed.Sindhi last updated on 22/Jun/20

T o T i n k u t a r a, t h e d e v e l o p e r

S i r,

I {d o n}^{'} t r e c e i v e n o t i c i c a t i o n a t t h e

m o m e n t .

Commented by Tinku Tara last updated on 23/Jun/20

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Commented by Rasheed.Sindhi last updated on 23/Jun/20

T h a n k s s i r

A f t e r t h i s c o m p l a i n t I r e c e i v e d

t h e n o t i f i c a t i o n o f y o u r r e p l y .

I t s e e m s t h a t t h e p r o b l e m w a s

t e m p o r a r y \dots . I u s e d o n l y o n e

d e v i c e .

Answered by maths mind last updated on 22/Jun/20

k x - 1 ⩽ E (k x) ⩽ k x

\Rightarrow

\underset{k = 1}{\sum^{n}} (k x - 1) ⩽ \underset{k = 1}{\sum^{n}} E (k x) ⩽ \underset{k = 1}{\sum^{n}} k x

\Leftrightarrow

\frac{n}{2} (n + 1) x - n ⩽ \underset{k = 1}{\sum^{n}} E (k x) ⩽ \frac{x}{2} n (n + 1)

\Rightarrow \frac{x}{2} + \frac{x}{2 n} - \frac{1}{n} ⩽ \frac{1}{n^{2}} E (k x) ⩽ \frac{x}{2} (1 + \frac{1}{n})

Commented by Ar Brandon last updated on 22/Jun/20

Thanks ��

Answered by maths mind last updated on 22/Jun/20

\lim_{n \to \infty} \underset{k = 0}{\sum^{n}} {(\begin{matrix} n \\ k \end{matrix})}^{-} ?

Commented by Ar Brandon last updated on 22/Jun/20

I think {it}^{'} s \overset{n}{} C_{k}

Answered by mathmax by abdo last updated on 22/Jun/20

a) S_{n} = \frac{1}{n^{2}} \sum_{k = 1}^{n} [kx] we have [kx] ⩽ kx < [kx] + 1 \Rightarrow kx - 1 < [kx] ⩽ kx

\Rightarrow \sum_{k = 1}^{n} (kx - 1) < \sum_{k = 1}^{n} [kx] ⩽ \sum_{k = 1}^{n} (kx) \Rightarrow \frac{xn (n + 1)}{2} - n < \sum_{k = 1}^{n} [kx] ⩽ \frac{xn (n + 1)}{2}

\frac{n (n + 1)}{{2 n}^{2}} x - \frac{1}{n} < \frac{1}{n^{2}} \sum_{k = 1}^{n} [kx] ⩽ \frac{n (n + 1)}{2 n} x we passe to limit we get

\lim_{n \to + \infty} S_{n} = \frac{x}{2}

Commented by Ar Brandon last updated on 22/Jun/20

Thank you Sir ��

Commented by mathmax by abdo last updated on 23/Jun/20

you are welcome