Use the first fundamental theorem of calculus to find?

Question

limit as n goes to infinity, sigma sign with 2n on top and i=n+1 on bottom, 1/i

Answer 1

The first thing to do is note that, whenever a and b are positive integers and b≥a, we have:

[k=a, b]∑1/k = [a, b+1]∫1/⌊x⌋ dx

Proof: note that since whenever n is an integer ⌊x⌋ = n almost everywhere on [n, n+1]. Thus, [n, n+1]∫1/⌊x⌋ dx = [n, n+1]∫1/n dx = 1/n. Thus we have [a, b+1]∫1/⌊x⌋ dx = [k=a, b]∑[k, k+1]∫1/⌊x⌋ dx = [k=a, b]∑1/k dx, as required.

Now note that since x-1 ≤ ⌊x⌋ ≤ x, we have 1/(x-1) ≥ 1/⌊x⌋ ≥ 1/x. Thus by monotonicity, we have that:

[n+1, 2n+1]∫1/(x-1) dx ≥ [n+1, 2n+1]∫1/⌊x⌋ dx ≥ [n+1, 2n+1]∫1/x dx.

Now, resolving these integrals, we have that:

[n+1, 2n+1]∫1/(x-1) dx = ln (2n) - ln n = ln 2
[n+1, 2n+1]∫1/⌊x⌋ dx = [k=n+1, 2n]∑1/k
[n+1, 2n+1]∫1/x dx = ln (2n+1) - ln (n+1) = ln ((2n+1)/(n+1))

Thus per the above inequality, we have:

ln 2 ≥ [k=n+1, 2n]∑1/k ≥ ln ((2n+1)/(n+1))

But [n→∞]lim ln 2 = [n→∞]lim ln ((2n+1)/(n+1)) = ln 2, so by the squeeze theorem, [n→∞]lim [k=n+1, 2n]∑1/k = ln 2. And we are done.