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加深影响!!! a=0, 1, 2, 3, 4, 5, 6, 7, 8, 9, .........
\(\displaystyle\lim_{n\to\infty}\bigg[\sum_{k=1}^n\frac{\sqrt{k(k+a)}}{n}-\frac{n}{2}\bigg]=\frac{a+1}{2}\)
因为我们恒有 a=0, 1, 2, 3, 4, 5, 6, 7, 8, 9, ... n=1, 2, 3, 4, 5, 6, 7, 8, 9, ...
\(\displaystyle\sum_{k=1}^n\frac{\sqrt{k(k+a)+(a/2)^2}}{n}-\frac{n}{2}≡\frac{a+1}{2}\) |
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