limn→∞∑r=115n n/(n2 + r2) is equal to:

limn→∞∑r=115n n/(n2 + r2)
limn→∞∑r=115n 1/(n + r2/n)
= limn→∞ 1/n ∑r=115n 1/(1 + (r/n)2)
This is a Riemann sum.
Let f(x) = 1/(1 + x2)
Then the above limit is equal to ∫015 f(x) dx = ∫015 1/(1 + x2) dx
= [tan⁻¹(x)]015 = tan⁻¹(15) - tan⁻¹(0) = tan⁻¹(15)
However, the question has a typographical error. It should read:
limn→∞∑r=1n n/(n2 + r2)
limn→∞∑r=1n 1/(n(1 + (r/n)2))
= limn→∞ (1/n)∑r=1n 1/(1 + (r/n)2)
This is the Riemann sum for ∫01 1/(1 + x2) dx
= [tan⁻¹(x)]01 = tan⁻¹(1) - tan⁻¹(0) = π/4