let-S-n-k-1-n-k-2-2k-1-2k-1-1-determine-S-n-interms-of-n-2-find-lim-n-S-n-

Question Number 48493 by maxmathsup by imad last updated on 24/Nov/18

l e t S_{n} = \sum_{k = 1}^{n} \frac{k^{2}}{(2 k - 1) (2 k + 1)}

1) d e t e r m i n e S_{n} i n t e r m s o f n

2) f i n d {l i m}_{n \to + \infty} S_{n}

Commented by maxmathsup by imad last updated on 25/Nov/18

1) f i r s t l e t d e c o m p o s e F (x) = \frac{x^{2}}{(2 x - 1) (2 x + 1)} \Rightarrow

F (x) = \frac{x}{4} (\frac{1}{2 x - 1} + \frac{1}{2 x + 1}) = \frac{1}{8} (\frac{2 x}{2 x - 1} + \frac{2 x}{2 x + 1})

= \frac{1}{8} (\frac{2 x - 1 + 1}{2 x - 1} + \frac{2 x + 1 - 1}{2 x + 1}) = \frac{1}{8} (1 + \frac{1}{2 x - 1} + 1 - \frac{1}{2 x + 1})

= \frac{1}{4} + \frac{1}{8} (\frac{1}{2 x - 1} - \frac{1}{2 x + 1}) \Rightarrow S_{n} = \frac{n}{4} + \frac{1}{8} \sum_{k = 1}^{n} \frac{1}{2 k - 1} - \frac{1}{8} \sum_{k = 1}^{n} \frac{1}{2 k + 1}

b u t \sum_{k = 1}^{n} \frac{1}{2 k + 1} =_{k + 1 = j} \sum_{j = 2}^{n + 1} \frac{1}{2 j - 1} = \sum_{j = 1}^{n} \frac{1}{2 j - 1} - 1 + \frac{1}{2 n + 1} \Rightarrow

S_{n} = \frac{n}{4} + \frac{1}{8} \sum_{k = 1}^{n} \frac{1}{2 k - 1} - \frac{1}{8} (\sum_{k = 1}^{n} \frac{1}{2 k - 1} - 1 + \frac{1}{2 n + 1})

= \frac{n}{4} + \frac{1}{8} - \frac{1}{8 (2 n + 1)}

2) i t s c l e a r t h a t {l i m}_{n \to + \infty} S_{n} = + \infty .

Answered by tanmay.chaudhury50@gmail.com last updated on 24/Nov/18

1) \frac{k^{2}}{(2 k + 1) (2 k - 1)}

= \frac{1}{4} [\frac{4 k^{2} - 1 + 1}{(2 k + 1) (2 k - 1)}]

= \frac{1}{4} [1 + \frac{1}{2} \times \frac{(2 k + 1) - (2 k - 1)}{(2 k + 1) (2 k - 1}]

T_{k} = \frac{1}{4} + \frac{1}{8} [\frac{1}{2 k - 1} - \frac{1}{2 k + 1}]

T_{1} = \frac{1}{4} + \frac{1}{8} [\frac{1}{1} - \frac{1}{3}]

T_{2} = \frac{1}{4} + \frac{1}{8} [\frac{1}{3} - \frac{1}{5}]

T_{3} = \frac{1}{4} + \frac{1}{8} [\frac{1}{5} - \frac{1}{7}]

\dots

\dots

T_{n} = \frac{1}{4} + \frac{1}{8} [\frac{1}{2 n - 1} - \frac{1}{2 n + 1}]

a d d t h e m

S_{n} . = \frac{n}{4} + \frac{1}{8} [1 - \frac{1}{2 n + 1}]

w h e n n \to \infty

S_{n} = \infty + \frac{1}{8} [1 - 0] = \infty

p l s c h e c k \dots

Commented by maxmathsup by imad last updated on 25/Nov/18

c o r r e c t a n s w e r t h a n k s .

Answered by ajfour last updated on 25/Nov/18

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

= \frac{1}{4} Σ \frac{k [(2 k + 1) + (2 k - 1)]}{(2 k - 1) (2 k + 1)}

4 S_{n} = Σ \frac{k}{2 k - 1} + Σ \frac{k}{2 k + 1}

8 S_{n} = Σ (1 + \frac{1}{2 k - 1}) + Σ (1 - \frac{1}{2 k + 1})

= 2 n + (\frac{1}{1} + \frac{1}{3} + . . + \frac{1}{2 n - 1})

- (\frac{1}{3} + \frac{1}{5} + \dots + \frac{1}{2 n + 1})

S_{n} = \frac{n}{4} + \frac{1}{8} (1 - \frac{1}{2 n + 1}) .

Commented by maxmathsup by imad last updated on 25/Nov/18

c o r r e c t a n s w e r t h a n k s