Question-130472

Question Number 130472 by shaker last updated on 26/Jan/21

Answered by mathmax by abdo last updated on 26/Jan/21

let w (x) = \sum_{k = 1}^{n} (2 k - 1) x^{k} \Rightarrow w (x) = 2 \sum_{k = 1}^{n} {kx}^{k} - \sum_{k = 1}^{n} x^{k}

we have \sum_{k = 0}^{n} x^{k} = \frac{x^{n + 1} - 1}{x - 1} (x \neq 1) \Rightarrow \sum_{k = 1}^{n} {kx}^{k - 1} = \frac{{nx}^{n + 1} - (n + 1) x^{n} + 1}{{(x - 1)}^{2}}

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

w (x) = \frac{2 x}{{(x - 1)}^{2}} ({nx}^{n + 1} - (n + 1) x^{n} + 1) - (\frac{1}{1 - x} - 1)

= \frac{2 x ({nx}^{n + 1} - (n + 1) x^{n} + 1)}{{(x - 1)}^{2}} - \frac{x (1 - x)}{{(1 - x)}^{2}}

= \frac{2 x ({nx}^{n + 1} - (n + 1) x^{n} + 1) + x^{2} - x}{{(x - 1)}^{2}} \Rightarrow

f (x) = \frac{\sqrt{{2 nx}^{n + 2} - 2 (n + 1) x^{n + 1} + x^{2} + x} \times \frac{1}{x - 1} - n}{(x - 1)}

= \frac{\sqrt{{2 nx}^{n + 2} - 2 (n + 1) x^{n + 1} + x^{2} + x} - n (x - 1)}{{(x - 1)}^{2}} we do the changement

x - 1 = t (t \to 0) \Rightarrow f (x) = f (t + 1)

= \frac{\sqrt{2 n {(t + 1)}^{n + 2} - 2 (n + 1) {(t + 1)}^{n + 1} + {(t + 1)}^{2} + t} - nt}{t^{2}}

we have {(1 + t)}^{n + 2} \sim 1 + (n + 2) t + \frac{(n + 2) (n + 1)}{2} t^{2}

{(1 + t)}^{n + 1} \sim 1 + (n + 1) t + \frac{(n + 1) n}{2} t^{2}

{(t + 1)}^{2} + t \sim 1 \Rightarrow

2 n {(t + 1)}^{n + 2} - 2 (n + 1) {(t + 1)}^{n + 1} + {(t + 1)}^{2} + t

\sim 2 n {1 + (n + 2) t + \frac{(n + 2) (n + 1)}{2} t^{2}} - 2 (n + 1) (1 + (n + 1) t + \frac{(n + 1) n}{2} t^{2}) + 1

\dots . be continued \dots

Answered by mathmax by abdo last updated on 26/Jan/21

in this case it better to use [hospital theorem let

u (x) = \sqrt{x + {3 x}^{2} + {5 x}^{3} + \dots . + (2 n - 1) x^{n}} - n and v (x) = x - 1

u^{^{'}} (x) = \frac{1 + 6 x + {15 x}^{2} + \dots . + n (2 n - 1) x^{n - 1}}{2 \sqrt{x + {3 x}^{2} + {5 x}^{3} + \dots . + (2 n - 1) x^{n}}} \Rightarrow

\lim_{x \to 1} u^{^{'}} (x) = \frac{1 + 6 + 15 + \dots . + n (2 n - 1)}{2 \sqrt{1 + 3 + 5 + \dots . + 2 n - 1}}

v^{^{'}} (x) = 1 \Rightarrow \lim_{x \to 1} v^{^{'}} (x) = 1 also

\sum_{k = 1}^{n} k (2 k - 1) = 2 \sum_{k = 1}^{n} k^{2} - \sum_{k = 1}^{n} k = 2 . \frac{n (n + 1) (2 n + 1)}{6} - \frac{n (n + 1)}{2}

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

= n (n + 1) (\frac{4 n + 2 - 3}{6}) = \frac{n (n + 1) (4 n - 1)}{6}

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

= \frac{n (n + 1)}{2} \Rightarrow \lim_{x \to 1} (\dots)

= \frac{\frac{n (n + 1) (4 n - 1)}{6}}{2 \sqrt{\frac{n (n + 1)}{2}}} = \frac{n (n + 1) (4 n - 1)}{6 \sqrt{2} \sqrt{n (n + 1)}}

= \frac{(4 n - 1) \sqrt{n (n + 1)}}{6 \sqrt{2}}