How can we compute the following limit: limn→∞(1+1n2)(1+2n2)…(1+nn2)

Question

How can we compute the following limit:  limn→∞(1+1n2)(1+2n2)…(1+nn2)

kovaje5w · Accepted Answer

We will estimate  f(n)=∑k=1nlog⁡(1+kn2)  Recall that log⁡(1+x)=x+O(x2). Hence, we get that  f(n)=∑k=1n(kn2+O(k2n4))=n+12n+O(1n) (1)  where we made use of the fact that ∑k=1nk=n(n+1)2 and ∑k=1nk2=n(n+1)(2n+1)6  Now letting n→∞ in (1), we get that  limn→∞f(n)=12  The product you are interested in is ef(n) and since ex is continuous, we have  limn→∞ef(n)=elimn→∞f(n)=e12=e

Deegan Mullen · Accepted Answer

For 1≤k≤n, we have  log⁡(1+kn2)−kn(n+1)=∫11+kn2dtt−kn(n+1)  =∫0kn2(11+t−nn+1)dt  =1n+1∫0kn2(1−tn1+t)dt  ≤1n+1∫0kn2dt  ≤kn2(n+1)  ≤1n(n+1)  Note, the integral in the third line is clearly non-negative, which means the initial expression is also non-negative. Therefore  0≤log⁡(1+kn2)−kn(n+1)≤1n(n+1)  Consequently, since 1+2+…+n=n(n+1)2, we have  0≤∑k=1nlog⁡(1+kn2)−12=∑k=1n(log⁡(1+kn2)−kn(n+1))≤1n+1  So by the Squeeze Theorem  limn→∞∑k=1nlog⁡(1+kn2)=12  and thus  limn→∞(1+1n2)(1+2n2)…(1+nn2)=e12=e

RizerMix · Accepted Answer

Using limn→∞(1+an)n=ea We have L=limn→∞[(1+1n2)(1+2n2)(1+3n2)...(1+nn2))n2]1n2 =limn→∞e1+...+nn2=limn→∞en(n+1)2n2=e

How can we compute the following limit: \lim_{n\to\infty}(1+1/n^2)(1+2/n^2)...(1+n/n^2)

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