advanced-calculus-lim-n-1-n-x-x-2-dx-n-1-solution-n-1-n-x-x-2-dx-k-1-n-1-k-k-1-x

Question Number 139556 by mnjuly1970 last updated on 28/Apr/21

\dots \dots . . a d v a n c e d \dots \dots \dots c a l c u l u s \dots \dots . .

Φ = {l i m}_{n \to \infty} {\int_{1}^{n} \frac{x}{{[x]}^{2}} d x - ψ (n + 1)} = ?

s o l u t i o n :

Φ_{n} = \int_{1}^{n} \frac{x}{{[x]}^{2}} d x = \underset{k = 1}{\sum^{n - 1}} \int_{k}^{k + 1} \frac{x}{k^{2}} d x

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

Φ = {l i m}_{n \to \infty} (Φ_{n} - ψ (n + 1))

= \frac{π^{2}}{12} + {l i m}_{n \to \infty} (\underset{k = 1}{\sum^{n - 1}} \frac{1}{k} - ψ (n + 1))

\underset{2 : ψ (n + 1) = H_{n} - γ}{\overset{1 : ψ (n + 1) := \frac{1}{n} + ψ (n)}{=}} \frac{π^{2}}{12} + {l i m}_{n \to \infty} (\underset{k = 1}{\sum^{n - 1}} \frac{1}{k} - H_{n} + γ)

∴ Φ := \frac{π^{2}}{12} + γ - {l i m}_{n \to \infty} (\frac{1}{n})

\dots \dots \dots Φ := \frac{1}{2} ζ (2) + γ

γ :: E u l e r - M a s c h e r o n i c o n s t a n t \dots