Here we have to check whether the series will converge or diverge. ∑ [n!(n+1)!]/(3n)!

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Arun · Answer

Dear student Just check two successive terms if next term is smaller than preceding term then it is a converging series. n = 1 1/3 nowput n = 2 12/6! = 1/60

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Here we have to check whether the series will converge or diverge. ∑ [n!(n+1)!]/(3n)!

Here we have to check whether the series will converge or diverge.
∑ [n!(n+1)!]/(3n)!

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Here we have to check whether the series will converge or diverge. ∑ [n!(n+1)!]/(3n)!

Here we have to check whether the series will converge or diverge.∑ [n!(n+1)!]/(3n)!

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Here we have to check whether the series will converge or diverge.
∑ [n!(n+1)!]/(3n)!