Final step on 'Sum of Gaps'

Calculus Level 5

Define a m + 1 , n = ∑ k = 1 n ( a m , k − b m ) and b m + 1 = lim ⁡ n → ∞ a m + 1 , n where m ≥ 0 \displaystyle a_{m+1,n} = \sum_{k=1}^{n} \left(a_{m,k} - b_m\right) \text{ and} \,\,\, b_{m+1} = \lim_{n \to \infty} a_{m+1,n} \text{where}\,\,\, m\ge 0

a 0 , k = ( − 1 ) k + 1 k and b 0 = 0 \displaystyle a_{0,k} = \frac{\left(-1\right)^{k+1}}{k} \text{ and}\,\,\, \displaystyle b_{0} = 0

It is well-known that b 1 = ln ⁡ 2 \displaystyle b_1 = \ln{2}

The value of b 2 b_2 and b 3 b_3 has been found at this problem and this problem respectively.

Now A = ∑ k = 1 ∞ b k \displaystyle A = \sum_{k=1}^{\infty} b_k .

What is the value of ⌊ 1000 A ⌋ , \displaystyle\left\lfloor 1000A \right\rfloor, where ⌊ ⋅ ⌋ \lfloor \cdot \rfloor denotes the floor function ?


The answer is 1000.

Chan Lye Lee by Chan Lye Lee , 44, Singapore

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1 solution

Mark Hennings
Nov 11, 2018

It is a simple induction to show that a m , n = ∫ 0 1 x m − 1 ( 1 − ( − x ) n ) ) ( 1 + x ) m d x b m = ∫ 0 1 x m − 1 ( 1 + x ) m d x a_{m,n} \; = \; \int_0^1 \frac{x^{m-1}\big(1 - (-x)^n)\big)}{(1+x)^{m}}\,dx \hspace{2cm} b_m \; = \; \int_0^1 \frac{x^{m-1}}{(1+x)^{m}}\,dx for all m , n ≥ 1 m,n \ge 1 , and hence A = ∑ m = 1 ∞ b m = ∑ m = 1 ∞ ∫ 0 1 x m − 1 ( 1 + x ) m d x = ∫ 0 1 d x = 1 A \; = \; \sum_{m=1}^\infty b_m \; = \; \sum_{m=1}^\infty \int_0^1 \frac{x^{m-1}}{(1+x)^m}\,dx \; = \; \int_0^1 \,dx \; = \; 1 making the answer 1000 \boxed{1000} .

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