Nature-and-Sum-u-n-where-u-n-0-1-x-n-1-1-x-x-n-dx-

Question Number 126776 by snipers237 last updated on 24/Dec/20

N a t u r e a n d S u m Σ u_{n}

w h e r e u_{n} = \int_{0}^{1} \frac{x^{n + 1}}{1 + x + \dots + x^{n}} d x

Answered by mindispower last updated on 24/Dec/20

1 + x + \dots \dots + x^{n} ⩾ (n + 1) . {(x^{\underset{k = 0}{\sum^{n}} k})}^{\frac{1}{n + 1}} = (n + 1) x^{\frac{n (n + 1)}{2 (n + 1)}} \dots . A M G

1 + x + \dots \dots + x^{n} ⩾ (n + 1) . {(x^{\underset{k = 0}{\sum^{n}} k})}^{\frac{1}{n + 1}} = (n + 1) x^{\frac{n (n + 1)}{2 (n + 1)}} \dots . A M G M

i n q u a l i t y

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

\Rightarrow U_{n} ⩽ \int_{0}^{1} \frac{x^{\frac{n + 2}{2}}}{n + 1} d x = \frac{2}{(n + 1) (n + 4)}

Σ U_{n} ⩽ 2 Σ \frac{1}{(n + 1) (n + 4)} \dots w i t c h c v

2 n d W a y

s t a r t e w i t h e

Ψ (z + 1) = - γ + \int_{0}^{1} \frac{1 - x^{z}}{1 - x} d x \dots . (I)

U_{n} = \int_{0}^{1} \frac{x^{n + 1}}{\underset{k = 0}{\sum^{n}} x^{k}} d x = \int_{0}^{1} \frac{x^{n + 1} (1 - x)}{1 - x^{n + 1}} d x

y = x^{n + 1} \Rightarrow d x = \frac{y^{- \frac{n}{n + 1}}}{n + 1} d y

\Leftrightarrow U_{n} = \frac{1}{n + 1} \int_{0}^{1} \frac{y (1 - y^{\frac{1}{n + 1}}) y^{- \frac{n}{n + 1}}}{1 - y} d y

U_{n} = \frac{1}{n + 1} \int_{0}^{1} \frac{y^{\frac{1}{n + 1}} - y^{\frac{2}{n + 1}}}{1 - y} d y

= \frac{1}{n + 1} [\int_{0}^{1} \frac{y^{\frac{1}{n + 1}} - 1}{1 - y} d y - \int_{0}^{1} \frac{y^{\frac{2}{n + 1}} - 1}{1 - y} d y]

B y a p p l y I

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

Ψ (1 + x) - Ψ (1 + y) = (y - x) Ψ^{1} (1 + c)

B y a p p l y m e a n v a l u e

t h e o r e m

\exists c \in [y, x]

\Rightarrow Ψ (1 + \frac{2}{n + 1}) - Ψ (1 + \frac{1}{n + 1}) = \frac{1}{n + 1} Ψ^{1} (1 + c)

c \in [\frac{1}{n + 1}; \frac{2}{n + 1}]

Ψ^{1} (z) = \sum_{n ⩾ 0} \frac{1}{{(n + z)}^{2}}

Ψ^{(1)} (1 + c) = \sum_{n ⩾ 0} \frac{1}{{(1 + c + n)}^{2}} ⩽ \sum_{n ⩾ 0} \frac{1}{{(1 + n)}^{2}} = ζ (2) = \frac{π^{2}}{6}

\Rightarrow U_{n} = \frac{Ψ (1 + \frac{2}{n + 1}) - Ψ (1 + \frac{1}{n + 1})}{n + 1}

Ψ (1 + \frac{2}{n + 1}) - Ψ (1 + \frac{1}{n + 1}) ⩽ \frac{1}{n + 1} . Ψ^{(1)} (1) = \frac{π^{2}}{6 (n + 1)}

\Rightarrow U_{n} ⩽ \frac{π^{2}}{6 {(n + 1)}^{2}} \Rightarrow Σ U_{n} ⩽ ζ^{2} (2) = \frac{π^{4}}{36}