lim _(n → ∞) [ 1/n + n/(n + 1)² + n/(n + 2)² + ........ + n/(2n + 1)² ] is equal to :

Solution

Maths sir · 4 steps

Write every term as n/(n+r)², pull out 1/n, and the sum becomes an integral from 0 to 1.

1. Common pattern. The first term fits the pattern too, because 1/n = n/(n+0)². So the bracket is Σᵣ₌₀ⁿ⁺¹ n/(n+r)².
2. Pull out 1/n. Divide the top and bottom of each term by n²: n/(n+r)² = 1/n · 1/(1+r/n)². This is a Riemann sum with r/n → x and 1/n → dx. The term for r = n+1 adds only one extra piece of width 1/n, which goes to 0.

   lim(n → ∞) 1/nΣᵣ₌₀ⁿ⁺¹ 1/(1+r/n)² = ∫₀¹ dx/(1+x)²

3. Find the limits. As r runs from 0 to about n, x = r/n runs from 0 to 1. The integral is [-1/(1+x)]₀¹ = -1/2 + 1 = 1/2.
4. So. The limit equals 1/2, which is option (a).

The idea to keep. To turn a sum into an integral, make each term look like 1/n f(r/n). The limits of x come from the first and last values of r/n. A few extra terms do not change the answer.

Answer from the NTA final answer key. Solution by PrepWiser.