This is the limit

Calculus Level 5

limnr=1nln(n2+r2)2ln(n)n=ln(2)+π22 \large \lim_{n\to\infty} \sum_{r=1}^n \frac{\ln(n^2+r^2) - 2\ln(n)}{n} = \ln(2) + \frac\pi2-2

We are given the value of the limit above. Suppose we consider the limit below

limn1n2m[(n2+1)(n2+22)(n2+32)(n2+n2)]mn \large \lim_{n\to\infty} \frac1{n^{2m}} \left[ (n^2+1)(n^2+2^2)(n^2+3^2)\ldots(n^2+n^2)\right]^{\frac mn}

If this limit equals to (aeba)m (ae^{b-a})^m for constant mm and positive integer aa, find the value of bb.

Clarification: e=limL0(1+L)1/L\displaystyle e= \lim_{L \to 0} (1+L)^{1/L } .


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