evaluate :-
lim (existing from n to infinity)1/n^2 [1/1+cos(pie/2n) +2/1+cos(2pie/2n) + 3/1+cos(3pie/2n)+......to n terms]
To evaluate the limit, we can use the concept of the Riemann sum, which can help approximate the limit as the number of terms approaches infinity.
Let's simplify the expression first:
lim (n -> infinity) 1/n^2 [1/(1 + cos(π/2n)) + 2/(1 + cos(2π/2n)) + 3/(1 + cos(3π/2n)) + ... + n/(1 + cos(nπ/2n))]
Now, let's rewrite the series using a common denominator:
lim (n -> infinity) 1/n^2 [(1 * (1 + cos(2π/2n)) * (1 + cos(3π/2n)) * ... * (1 + cos(nπ/2n)) +
2 * (1 + cos(π/2n)) * (1 + cos(3π/2n)) * ... * (1 + cos(nπ/2n)) +
3 * (1 + cos(π/2n)) * (1 + cos(2π/2n)) * ... * (1 + cos(nπ/2n)) +
... +
n * (1 + cos(π/2n)) * (1 + cos(2π/2n)) * ... * (1 + cos((n-1)π/2n)))]
Now, let's take the limit of each term individually.
For the first term, as n approaches infinity, cos(2π/2n), cos(3π/2n), ... , cos(nπ/2n) all approach 1 since the argument approaches zero. Therefore, in the numerator, we have (1 * 1 * 1 * ... * 1) = 1. The denominator is n^2. So, the contribution of the first term in the limit is 1/n^2.
In a similar manner, we can evaluate the limit for all the other terms. Each term will have factors with arguments that approach zero, leading to the numerator being equal to its corresponding term number, and the denominator being n^2. So, the contribution of each term in the limit is the term number divided by n^2.
Therefore, the overall limit is:
lim (n -> infinity) 1/n^2 [1 + 2 + 3 + ... + n] = lim (n -> infinity) (n(n+1)/2n^2) = 1/2
Thus, the given limit is 1/2.