∞∑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: 1√1−4x=∞∑n=0(2nn)xn
Integrate both the sides under the limit 0 to x i.e −12(√1−4x−1)=∞∑n=0(2nn)xn+1n+1⇒−12x(√1−4x−1)=∞∑n=0(2nn)xnn+1
Substitute x=1/4 to get: ∞∑n=0(2nn)n+114n=2⇒∞∑n=1(2nn)n+114n=1
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