Interesting Limit

Calculus Level 4

lim n ( 1000 n + 1 + 1000 n + 2 + + 1000 n + n ) = ? \left \lfloor \lim_{n\to \infty } \left ( \frac{1000}{n+1}+\frac{1000}{n+2}+\cdots +\frac{1000}{n+n} \right) \right \rfloor = \, ?


The answer is 693.

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1 solution

lim n 1000 ( 1 n + 1 + 1 n + 2 + + 1 n + n ) = lim n 1000 n r = 1 n 1 1 + r n \lim_{n\rightarrow{∞}} 1000(\dfrac{1}{n+1}+\dfrac{1}{n+2}+\cdots+\dfrac{1}{n+n})= \lim_{n\rightarrow{∞}}\dfrac{1000}{n} \sum_{r=1}^n \dfrac{1}{1+\dfrac{r}{n}} Applying Riemann sums It is 1000 0 1 1 1 + x d x = [ 1000 log ( 1 + x ) ] 0 1 1000\int_0^1 \dfrac{1}{1+x} dx = [1000 \log(1+x)]_0^1 = 1000 log ( 2 ) = 693.147 = 1000\log(2)= 693.147\cdots Answer is 1000 log ( 2 ) = 693 ⌊1000\log(2)⌋=693

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