let-put-s-n-k-0-k-n-2k-1-k-1-k-n-k-find-lim-n-gt-s-n-

Question Number 26761 by abdo imad last updated on 29/Dec/17

l e t p u t s_{n} = \frac{\sum_{k = 0}^{k = n} (2 k + 1)}{\sum_{k = 1}^{k = n} k}

f i n d {l i m}_{n - >\propto} s_{n}

Answered by prakash jain last updated on 29/Dec/17

s_{n} = \frac{\sum_{k = 0}^{k = n} (2 k + 1)}{\sum_{k = 1}^{k = n} k} = \frac{{(n + 1)}^{2}}{n (n + 1) / 2} = \frac{2 (n + 1)}{n}

\lim_{n \to \infty} \frac{2 (n + 1)}{n} = 2

Commented by abdo imad last updated on 29/Dec/17

t h e s e q u e n c e v_{n} = 2 n + 1 i s a r i t h m e t i c s o

\sum_{k = 0}^{k = n} (2 k + 1) = v_{0} + v_{1} + \dots v_{n} = \frac{n + 1}{2} (v_{0} + v_{n}) = \frac{n + 1}{2} (1 + 2 n + 1)

= {(n + 1)}^{2} ? a n d f o r t h e s a m e r a i s o n \sum_{k = 1}^{k = n} k = \frac{n (n + 1)}{2}

{l i m}_{n - >\propto} s_{n} = {l i m}_{n - >\propto} 2 \frac{{(n + 1)}^{2}}{n^{2} + n} = 2

Commented by prakash jain last updated on 29/Dec/17

Thanks . I corrected .

Earlier

\underset{i = 0}{\sum^{n}} (2 i + 1) = n^{2} (i t s h o u l d h a v e b e e n

{(n + 1)}^{2} .