Capital and lowercase gamma

Calculus Level 3

L = lim ⁡ x → 0 ( 2015 x ) ! − 1 x \large{ L=\displaystyle \lim_{x \rightarrow 0} \dfrac{(2015x)!-1}{x} }

If L L can be represented in the form − A ⋅ γ -A \cdot \gamma , where γ \gamma is the Euler-Mascheroni constant, find the value of A A .


The answer is 2015.

Hobart Pao by Hobart Pao , 23, USA

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1 solution

Surya Prakash
Sep 27, 2015

lim ⁡ x → 0 ( 2015 x ) ! − 1 x = lim ⁡ x → 0 Γ ( 2015 x + 1 ) − 1 x \lim_{x \rightarrow 0} \dfrac{(2015x)! - 1}{x} = \lim_{x \rightarrow 0} \dfrac{\Gamma (2015x+1) - 1}{x} By L' Hospital Rule,

lim ⁡ x → 0 2015 × Γ ′ ( 2015 x + 1 ) 1 = 2015 × ( lim ⁡ x → 0 Ψ ( 2015 x + 1 ) × Γ ( 2015 x + 1 ) ) \lim_{x \rightarrow 0} \dfrac{2015\times \Gamma ' (2015x+1)}{1} = 2015 \times \left( \lim_{x \rightarrow 0} \Psi (2015x + 1) \times \Gamma (2015x+1) \right)

= 2015 × Γ ( 1 ) × Ψ ( 1 ) = − 2015 γ =2015 \times \Gamma (1) \times \Psi (1) = -2015 \gamma

Therefore, A = 2015 A=\boxed{2015}

Note: Ψ ( x ) = Γ ′ ( x ) Γ ( x ) \Psi (x) = \dfrac{\Gamma ' (x)}{\Gamma (x)} . So, Γ ′ ( x ) = Ψ ( x ) × Γ ( x ) \Gamma ' (x) = \Psi (x) \times \Gamma (x) .

4 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

This may seem immediate, but there is a bunch of work involved in showing that these equations are true, other than accepting them as fact.

Even I used the digamma function.

Aditya Kumar - 5 years, 8 months ago

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