A calculus problem by Hamza A

Calculus Level 5

lim ⁡ n → ∞ ( a n γ γ a n ) 2 n \large \displaystyle\lim _{ n\to\infty }{ \left( \dfrac { { a }_{ n }^{ \gamma } }{ { \gamma }^{ { a }_{ n } } } \right) ^{ 2n } }

Let a n = H n − ln ⁡ n { a }_{ n }={ H }_{ n }-\ln { n } , where H n { H }_{ n } denote the n th n^{\text{th}} harmonic number .

If the limit above is equal to A e B γ \dfrac { Ae }{ B\gamma } for positive coprime integers A A and B B , find A + B A+B .

Notation : γ \gamma denote the Euler-Mascheroni constant , γ ≈ 0.5772 \gamma \approx 0.5772 .


The answer is 2.

Hamza A by Hamza A , 19, USA

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1 solution

Rohan Shinde
Dec 2, 2019

Note that as n → ∞ \displaystyle n\to\infty H n ∼ ln ⁡ ( n ) + γ + 1 2 n H_n\sim \ln(n)+\gamma+\frac{1}{2n}

Hence ξ = lim ⁡ n → ∞ ( a n γ γ a n ) 2 n \xi=\lim_{n\to\infty}\left(\frac{a_{n}^{\gamma}}{\gamma^{a_n}}\right)^{2n} = lim ⁡ n → ∞ ( ( γ + 1 2 n ) γ γ γ + 1 2 n ) 2 n =\displaystyle \lim_{n\to\infty}\left(\frac{\left(\gamma+\frac{1}{2n}\right)^{\gamma}}{\gamma^{\gamma+\frac{1}{2n}}}\right)^{2n} = lim ⁡ n → ∞ ( ( 1 + 1 2 n γ ) γ γ 1 2 n ) 2 n =\displaystyle \lim_{n\to\infty}\left(\frac{\left(1+\frac{1}{2n\gamma}\right)^{\gamma}}{\gamma^{\frac{1}{2n}}}\right)^{2n} = lim ⁡ n → ∞ ( 1 + 1 2 n γ ) 2 n γ γ =\displaystyle \lim_{n\to\infty}\frac{\left(1+\frac{1}{2n\gamma}\right)^{2n\gamma}}{\gamma}

But lim ⁡ n → ∞ ( 1 + 1 2 n γ ) 2 n γ = e \displaystyle \lim_{n\to\infty}\left(1+\frac{1}{2n\gamma}\right)^{2n\gamma}=e

Hence ξ = e γ \xi=\frac{e}{\gamma}

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