Sunday, December 4, 2016

sequences and series - Find the sum, if exists sumlimitsinftyn=1frac(2n)!22n(n!)2(n+1)


n=1(2n)!22n(n!)2(n+1)


By comparison test this series converges. Any nice way to work the sum?



I see that this can be written as: n=1(2nn)22n(n+1)


Answer




Consider the following expansion: 114x=n=0(2nn)xn

Integrate both the sides under the limit 0 to x i.e 12(14x1)=n=0(2nn)xn+1n+112x(14x1)=n=0(2nn)xnn+1
Substitute x=1/4 to get: n=0(2nn)n+114n=2n=1(2nn)n+114n=1


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