ANSWER: 4π
Explanation:
1+n2n+4+n2n+9+n2n+⋯+n2+n2n=n1(n21+ 11+n2 4+ 11+n29+ 11+⋯+n2 n2 + 11)= ∑k=1nn1⋅(nk)2+11 .
For the function f(x)=1+x21 this sum is the the integral Riemann sum on the interval [0,1]. Those sum ∑k=1nf(ck)(xk−xk−1) , where x0=0,xk=ck=nk, k=1,2,...,n . Since the function is continuous , it is integrable. Therefore, limn→∞(1+n2n+4+n2n+9+n2n+⋯+n2+n2n) =
=∫011+x2dx= arctan1−arctan0=4π .
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