How to compute ∏n=2∞n3−1n3+1

Question

How to compute  ∏n=2∞n3−1n3+1

Addison Kirk · Accepted Answer

HINT:  If  tn=n3−1n3+1=(n−1)(n2+n+1)(n+1)(n2−n+1)  tn+1=n[(n+1)2+n+1+1](n+2)[(n+1)2−(n+1)−1]=n[(n+1)2+n+1+1](n+2)(n2+n+1)  and  tn−1=(n−2)[(n−1)2+n−1+1]n[(n−1)2−(n−1)−1]=(n−2)(n2−n+1)n[(n−1)2−(n−1)−1]  Alternatively, let un=n−1n+1, vn=n2+n+1n2−n+1 so that tn=un⋅vn  ⇒∏2≤n≤run=1⋅2⋅3⋅…(r−2)(r−1)3⋅4⋅5⋅…r(r+1)2r(r+1)  ⇒∏2≤n≤rtn=(∏2≤n≤run)(∏2≤n≤rvn)=2(r2+r+1)3r(r+1)  Setting r→∞, 2(r2+r+1)3r

Maritza Mccall · Accepted Answer

Note that  n3−1n3+1=n−1n+1n2+n+1n2−n+1  Also note that  (n−1)2+(n−1)+1=n2−n+1  So the infinite product in question is really the product of two telescoping products.

RizerMix · Accepted Answer

r3−a3r3+a3=r−ar+a⋅r2+ra+a2r2−ra+a2 Let f(r)=r2−ra+a2 f(r+h)=(r+h)2−(r+h)a+a2=r2+r(2h−a)+a2−ah If we set r2+ra+a2=f(r+h) r2+ra+a2=r2+r(2h−a)+a2−ah Comparing the coefficients of r, a=2h−a→h=a Compare the constants a2=a2−ah+h2→h(h−a)=0 →h=a r2+ra+a2=f(r+a) r3−a3r3+a3=f(r+a)f(r)⋅g(r−2a)g(r) where g(m)=m+a Clearly ∏r=1nf(r+a)f(r)⋅g(r−2a)g(r) telescopes for finite integer a

How to compute \prod_{n=2}^\infty\frac{n^3-1}{n^3+1}

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