# Calc6.73

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 ${\frac {a}{b}}$.