Question-130181

Question Number 130181 by Adel last updated on 23/Jan/21

Answered by Olaf last updated on 23/Jan/21

\lim_{x \to 3} \frac{Γ (x + 1) - 6}{x x - 33}

= \lim_{x \to 3} \frac{Γ (x + 1) - 6}{11 (x - 3)}

= \lim_{x \to 3} \frac{Γ^{'} (x + 1)}{11}

= \lim_{x \to 3} \frac{Γ (x + 1) ψ (x + 1)}{11}

= \frac{Γ (4) ψ (4)}{11}

= \frac{3! (H_{3} - γ)}{11}

= \frac{6 (\frac{11}{6} - γ)}{11}

= 1 - \frac{6}{11} γ (1)

ψ (3) = H_{2} - γ = \frac{3}{2} - γ (2)

S = \underset{n = 1}{\sum^{\infty}} \frac{{(\frac{1}{2})}^{n}}{n}

\frac{1}{1 - x} = \underset{n = 0}{\sum^{\infty}} x^{n}

- \ln (1 - x) = \underset{n = 0}{\sum^{\infty}} \frac{x^{n + 1}}{n + 1} = \underset{n = 1}{\sum^{\infty}} \frac{x^{n}}{n}

For x = \frac{1}{2}, S = \ln 2 (3)

With (1), (2) and (3), the result is :

{{(1 - \frac{6}{11} γ)}^{\frac{3}{2} - γ}}^{\ln 2}

\approx {{(1 - \frac{6}{11} 0, 5772156649)}^{\frac{3}{2} - 0, 5772156649}}^{\ln 2}

\approx 0, 7851749027

Commented by Adel last updated on 23/Jan/21

thanks sir

Commented by 0731619177 last updated on 24/Jan/21

h h h h h h h

څه وایی تره بچه