nice-calculus-evaluate-lim-x-0-1-x-ln-1-x-x-x-1-

Question Number 124548 by mnjuly1970 last updated on 04/Dec/20

\dots n i c e c a l c u l u s \dots

e v a l u a t e :::

{l i m}_{x \to 0} {\frac{1}{x} [\frac{l n (Γ (1 + x)}{x} - ψ (x + 1)]} = ?

Answered by mindispower last updated on 04/Dec/20

l n (Γ (1 + x)) = f (x)

f (x) = f (0) + f^{'} (0) x + \frac{f^{″} (0) x^{2}}{2} + o (x^{2})

f (0) = 0, f^{'} (0) = Ψ (1), f^{″} (0) = Ψ^{1} (1)

Ψ (x + 1) = Ψ (1) + Ψ^{1} (1) x + o (x)

\frac{l n (Γ (1 + x))}{x} - Ψ (x + 1) = \frac{x Ψ (1) + \frac{x^{2}}{2} Ψ^{1} (1) + o (x^{2})}{x} - Ψ (1) - Ψ^{1} (1) x + o (x)

= - \frac{Ψ^{1} (1)}{2} x + o (x)

\lim_{x \to 0} {\frac{1}{x} [\frac{l n (Γ (1 + x))}{x} - Ψ (1 + x)]} = \lim_{x \to 0} \frac{1}{x} . \frac{- Ψ (1)}{2} x + o (x)

= \lim_{x \to 0} - \frac{Ψ^{1} (1)}{2} + o (1) = - \frac{Ψ^{1} (1)}{2}

Ψ^{1} (z) = \sum_{j ⩾ 0} \frac{1}{{(j + z)}^{2}} \Rightarrow Ψ^{1} (1) = \sum_{j ⩾ 0} \frac{1}{{(1 + j)}^{2}} = \frac{π^{2}}{6}

w e g e t - \frac{π^{2}}{12}

Commented by mnjuly1970 last updated on 04/Dec/20

b r a v o m r m i n d s p o w e r

m a c l a u r e n e x p a n s i o n o f Γ (x + 1)

g r a t e f u l \dots