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3 solutions
L=(1÷n) integration(0to1)(1÷(1+x))=(1÷n) ln(2)=0
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The definition of the Riemann Sum includes the n 1 in the integration, and here's why:
What we are essentially doing is breaking up the region into n rectangles each with width n 1 and the height of the m t h region is f ( n m ) , and then adding up their areas.
This gives us:
m = 1 ∑ n base × height = m = 1 ∑ n n 1 × 1 + n m 1 .
Letting the number of rectangles we use approach infinity gives us an increasingly accurate estimate, with the limit approaching the real area. This becomes analogous to the integral. Hope this helps!
I've found a simpler, but probably less elegant, solution. The first step is an easy rewriting of the sum: lim n → ∞ m = 1 ∑ n m + n 1 = n → ∞ lim ( k = 1 ∑ 2 n k 1 − k = 1 ∑ n k 1 )
Then the terms can be rearrenged properly: 1 + 1 / 2 − 1 + 1 / 3 + 1 / 4 − 1 / 2 + 1 / 5 + 1 / 6 − 1 / 3 . . . = 1 − 1 / 2 + 1 / 3 − 1 / 4 + 1 / 5 − 1 / 6 . . .
Which leads to the well-known result: n = 1 ∑ ∞ n ( − 1 ) n + 1 = ln ( 2 )
Relevant wiki: Riemann Sums
L = n → ∞ lim m = 1 ∑ n m + n 1 = n → ∞ lim m = 1 ∑ n n 1 1 + n m 1
This is a Riemann sum of the function 1 + x 1 , so we end up with:
L = ∫ 0 1 1 + x 1 d x = ln ( 2 ) − ln ( 1 ) = ln ( 2 )