Here we have the series Sn as,
Sn=n=1∑∞2nn(2n)!
(a)
Now, let an= 2nn(2n)!
So, an+1=2n+1(n+1)(2n+2)!
So, r=anan+1
r=anan+1 ⇒r=2n+1⋅(n+1)⋅(2n)!(2n+2)!⋅n⋅2n ⇒r=n(n+1)
Therefore, r=n(n+1)
(b)
Now, we will evaluate n→∞limr
So,
n→∞limr=n→∞limn(n+1)=∞
So, as r tends to ∞ ,we can conclude that the series Sn diverges.
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