find-approximate-value-of-3-by-using-n-1-n-n-1-for-n-integr-natural-

Question Number 57647 by maxmathsup by imad last updated on 09/Apr/19

f i n d a p p r o x i m a t e v a l u e o f ξ (3) b y u s i n g n - 1 ⩽ n ⩽ n + 1 f o r n i n t e g r

n a t u r a l .

Commented by maxmathsup by imad last updated on 18/Apr/19

l e t d e t e r m i n e S_{2} = \sum_{n = 2}^{\infty} \frac{1}{(n - 1) n^{2}} + 1 \Rightarrow S_{2} = {l i m}_{n \to + \infty} \sum_{k = 1}^{n} \frac{1}{k {(k + 1)}^{2}} + 1

l e t d e c o m p o s e G (x) = \frac{1}{x {(x + 1)}^{2}} \Rightarrow G (x) = \frac{a}{x} + \frac{b}{x + 1} + \frac{c}{{(x + 1)}^{2}}

a = {l i m}_{x \to + \infty} x G (x) = 1, c = {l i m}_{x \to - 1} {(x + 1)}^{2} G (x) = - 1 \Rightarrow

G (x) = \frac{1}{x} + \frac{b}{x + 1} - \frac{1}{{(x + 1)}^{2}}

G (1) = \frac{1}{4} = 1 + \frac{b}{2} - \frac{1}{4} \Rightarrow 1 = 4 + 2 b - 1 = 3 + 2 b \Rightarrow 2 b = - 2 \Rightarrow b = - 1 \Rightarrow

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

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

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

\sum_{k = 1}^{n} \frac{1}{k} = H_{n}, \sum_{k = 2}^{n + 1} \frac{1}{k} = H_{n + 1} - 1 a n d \sum_{k = 2}^{n + 1} \frac{1}{k^{2}} = ξ_{n + 1} (2) - 1 \Rightarrow

S_{2} = H_{n} - H_{n + 1} + 1 - ξ_{n + 1} (2) + 1 \to 2 - \frac{π^{2}}{6} \Rightarrow \frac{π^{2}}{6} - \frac{1}{2} ⩽ ξ (3) ⩽ 2 - \frac{π^{2}}{6}

\Rightarrow \frac{π^{2} - 3}{6} ⩽ ξ (3) ⩽ \frac{12 - π^{2}}{6}

Commented by maxmathsup by imad last updated on 18/Apr/19

w e h a v e (n - 1) n^{2} ⩽ n^{3} ⩽ (n + 1) n^{2} \Rightarrow \frac{1}{(n + 1) n^{2}} ⩽ \frac{1}{n^{3}} ⩽ \frac{1}{(n - 1) n^{2}} \Rightarrow

\sum_{n = 2}^{\infty} \frac{1}{(n + 1) n^{2}} ⩽ \sum_{n = 2}^{\infty} \frac{1}{n^{3}} ⩽ \sum_{n = 2}^{\infty} \frac{1}{(n - 1) n^{2}} \Rightarrow \sum_{n = 2}^{\infty} \frac{1}{(n + 1) n^{2}} ⩽ ξ (3) - 1

⩽ \sum_{n = 2}^{\infty} \frac{1}{(n - 1) n^{2}} \Rightarrow \sum_{n = 2}^{\infty} \frac{1}{(n + 1) n^{2}} + 1 ⩽ ξ (3) ⩽ \sum_{n = 2}^{\infty} \frac{1}{(n - 1) n^{2}} + 1

l e t d e t e r m i n e S_{1} = 1 + \sum_{n = 2}^{\infty} \frac{1}{(n + 1) n^{2}} \Rightarrow S_{1} = \frac{1}{2} + \sum_{n = 1}^{\infty} \frac{1}{(n + 1) n^{2}}

S_{1} = {l i m}_{n \to + \infty} S_{n} w i t h S_{n} = \frac{1}{2} + \sum_{k = 1}^{n} \frac{1}{(k + 1) k^{2}} l e t d e c o m p o s e F (x) = \frac{1}{(x + 1) x^{2}}

F (x) = \frac{a}{x + 1} + \frac{b}{x} + \frac{c}{x^{2}}

a = {l i m}_{x \to - 1} (x + 1) F (x) = 1, c = {l i m}_{x \to 0} x^{2} F (x) = 1 \Rightarrow F (x) = \frac{1}{x + 1} + \frac{b}{x} + \frac{1}{x^{2}}

F (1) = \frac{1}{2} = \frac{1}{2} + b + 1 \Rightarrow b = - 1 \Rightarrow F (x) = \frac{1}{x + 1} - \frac{1}{x} + \frac{1}{x^{2}} \Rightarrow

\sum_{k = 1}^{n} F (k) = \sum_{k = 1}^{n} \frac{1}{k + 1} - \sum_{k = 1}^{n} \frac{1}{k} + \sum_{k = 1}^{n} \frac{1}{k^{2}}

= \sum_{k = 2}^{n + 1} \frac{1}{k} - \sum_{k = 1}^{n} \frac{1}{k} + \sum_{k = 1}^{n} \frac{1}{k^{2}} = - 1 - \frac{1}{n + 1} + \sum_{k = 1}^{n} \frac{1}{k^{2}} \to \frac{π^{2}}{6} - 1 \Rightarrow

S_{1} = \frac{π^{2}}{6} - \frac{1}{2}