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Discuss the convergence or divergence of the series \sum _{{n=1}}^{{\infty }}{\frac  {n!}{(n+1)^{n}}}\left({\frac  {237}{50}}\right)^{n}.

This series looks tricky. Instinctively, since we have (n+1)^{n} on the bottom, we should think this series is going to converge, even with the n! in the numerator. But, let us use the root test to find out for sure.

\lim _{{n\rightarrow \infty }}{\sqrt[ {n}]{{\frac  {n!}{(n+1)^{n}}}\left({\frac  {237}{50}}\right)^{n}}}={\frac  {237}{50}}\lim _{{n\rightarrow \infty }}{\sqrt[ {n}]{{\frac  {n!}{(n+1)^{n}}}}}={\frac  {237}{50}}\lim _{{n\rightarrow \infty }}{\sqrt[ {n}]{{\frac  {1}{n+1}}\times {\frac  {2}{n+1}}\times {\frac  {3}{n+1}}\times \ldots \times {\frac  {n}{n+1}}}}

This limit looks to be 0, and in fact it is, but how can we show that. The trick is that as n goes to infinity, it can be arbitrarily large. We know that

{\frac  {237}{50}}\lim _{{n\rightarrow \infty }}{\sqrt[ {n}]{{\frac  {1}{n+1}}\times {\frac  {2}{n+1}}\times {\frac  {3}{n+1}}\times \ldots \times {\frac  {n}{n+1}}}}<{\frac  {237}{50}}K\lim _{{n\rightarrow \infty }}{\sqrt[ {n}]{{\frac  {n}{n+1}}\times {\frac  {n}{n+1}}\times {\frac  {n}{n+1}}\times \cdots \times {\frac  {n}{n+1}}}}

The thing here is that I can make K as small as I want. For here, making it {\frac  {1}{100}} suffices. Since we are taking the limit as n goes to infinity, we can assume n>10,000. If this is true, then in changing {\sqrt  {{\frac  {1}{n+1}}}} to {\sqrt  {{\frac  {n}{n+1}}}}, we multiplied by at least {\sqrt  {10,000}}. So, we can bring a {\frac  {1}{{\sqrt  {10,000}}}}={\frac  {1}{100}} outside of the square root, though we can bring out an arbitrarily small number by assuming n is as big as we choose. Thus, we have

{\frac  {237}{50}}\lim _{{n\rightarrow \infty }}{\sqrt[ {n}]{{\frac  {1}{n+1}}\times {\frac  {2}{n+1}}\times {\frac  {3}{n+1}}\times \ldots \times {\frac  {n}{n+1}}}}<{\frac  {237}{50}}\times {\frac  {1}{100}}{\sqrt[ {n}]{\left({\frac  {n}{n+1}}\right)^{n}}}={\frac  {237}{5000}}\lim _{{n\rightarrow \infty }}{\frac  {n}{n+1}}={\frac  {237}{5000}}<1

Thus, this series converges by the root test. Also, note that the series

\sum _{{n=1}}^{{\infty }}{\frac  {n!}{(n+1)^{n}}}\left({\frac  {a}{b}}\right)^{n}

converges which we can prove easily by assuming n is even bigger and pulling out an even smaller constant to make up for any {\frac  {a}{b}}.

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