Convergence of $ \sum _ { n = 1 } ^ { \infty } \frac { n ! n ^ { n } } { ( 2 n ) ! } $

Question

The task is to find out if this series is convergent or divergent.

$$ \sum _ { n = 1 } ^ { \infty } \frac { n ! n ^ { n } } { ( 2 n ) ! } $$

The solution uses the ratio test and says:

$ \left.\begin{aligned} \frac { a _ { n + 1 } } { a _ { n } } & = \frac { ( n + 1 ) ! ( n + 1 ) ^ { n + 1 } ( 2 n ) ! } { ( 2 ( n + 1 ) ) ! n ! n ^ { n } } = \frac { ( n + 1 ) n ! ( n + 1 ) ( n + 1 ) ^ { n } ( 2 n ) ! } { ( 2 n + 2 ) ( 2 n + 1 ) ( 2 n ) ! n ! n ^ { n } } \\ & = \frac { ( n + 1 ) ( n + 1 ) } { ( 2 n + 2 ) ( 2 n + 1 ) } \cdot \left( \frac { n + 1 } { n } \right) ^ { n } = \frac { \left( 1 + \frac { 1 } { n } \right) \left( 1 + \frac { 1 } { n } \right) } { \left( 2 + \frac { 2 } { n } \right) \left( 2 + \frac { 1 } { n } \right) } \cdot \left( \frac { n + 1 } { n } \right) ^ { n } \\ & \rightarrow \frac { 1 } { 4 } \cdot e < 1 \text { for } n \rightarrow \infty \end{aligned} \right. $

I understand every step until here

$$ \frac { \left( 1 + \frac { 1 } { n } \right) \left( 1 + \frac { 1 } { n } \right) } { \left( 2 + \frac { 2 } { n } \right) \left( 2 + \frac { 1 } { n } \right) } \cdot \left( \frac { n + 1 } { n } \right) ^ { n } $$

How can all n's on the left site become 1/n? And I understand how the left site can become $\frac{1}{4}$, but how can the right site become e in the last step?

Answer 1

$$\frac{(n+1)(n+1)}{(2n+2)(2n+1)}\cdot\left(\frac{n+1}{n}\right)^n$$

Divide both denominator and numerator by $n^2 $

$$\frac{\frac{(n+1)}{n}\frac{(n+1)}{n}}{\frac{(2n+2)}{n}\frac{(2n+1)}{n}}\cdot\left(\frac{n+1}{n}\right)^n$$ $$\frac{(1+\frac{1}{n})(1+\frac{1}{n})}{(2+\frac{2}{n})(2+\frac{1}{n})}\cdot\left(1+\frac{1}{n}\right)^n$$ $$\lim_{n \to \infty}\frac{(1+\frac{1}{n})(1+\frac{1}{n})}{(2+\frac{2}{n})(2+\frac{1}{n})}=\frac{1\cdot 1}{2 \cdot 2}=\frac{1}{4}$$ and$$\lim_{n \to \infty} (1+\frac{1}{n})^n=e$$

Answer 2

Hint: Multiply by $$ \frac{1/n^2}{1/n^2}.$$

Answer 3

We have that

$$\left( \frac { n + 1 } { n } \right) ^ { n }=\left( 1+\frac { 1 } { n } \right) ^ { n }\to e$$

and from here

$$\frac { ( n + 1 ) ( n + 1 ) } { ( 2 n + 2 ) ( 2 n + 1 ) } $$

just divide both numerator and denominator by $n^2$.

Answer 4

How can all n's on the left site become 1/n?

This is a standard method when finding the limit of such sequences. You factor out the highest power of $n$ to cancel it out.

So if we would like to involve one more step befor the one you do not follow it would be:

$\frac{(n+1)(n+1)}{(2n+2)(2n+1)}\cdot\left(\frac{n+1}{n}\right)^n=\frac{n^2(1+1/n)(1+1/n)}{n^2(2+2/n)(2+1/n)}\cdot\left(\frac{n+1}{n}\right)^n$

Also note, that $\lim_{n\to\infty} \left(\frac{n+1}{n}\right)^n=\lim_{n\to\infty} \left(1+\frac{1}{n}\right)^n=e$.

Answer 5

Divide both the numerator and denominator on the left side by $n^2$.

As n approaches $\infty$, the 1/n and 2/n tend to zero, giving $\frac{(1)(1)}{(2)(2)}$ = $\frac{1}{4}$.

The limit $\lim_{n\to\infty}(\frac{n+1}{n})^n$ = $\lim_{n\to\infty}(1+\frac{1}{n})^n$ = e.

Answer 6

$\sum _ { n = 1 } ^ { \infty } \frac { n ! n ^ { n } } { ( 2 n ) ! } $

For region of convergence, just replace $n!$ with $(n/e)^n$ and then use the n-th root test. The additional $\sqrt{n}$ only affects the boundaries.

In this case (why the annoying spaces?), $\dfrac{n!n^{n}}{(2n)!} $ becomes $\dfrac{(n/e)^nn^{n}}{(2n/e)^{2n}} =\left(\dfrac{(n/e)n}{(2n/e)^{2}}\right)^n =\left(\dfrac{n^2e^2}{4n^2e}\right)^n =\left(\dfrac{e}{4}\right)^n $

Since $\dfrac{e}{4} < 1$, the sum converges.

Answer 7

In order to find an explicit value for the given series, we may recall that for any $|x|<\frac{1}{e}$ we have $$ \sum_{n\geq 1}\frac{n^{n-1}}{n!} x^n = -W(-x) $$ with $W(x)$ being the inverse function of $x e^x$ (Lambert's function), by Lagrange inversion theorem.
A rapid sequence of formal manipulations:

$$ \sum_{n\geq 1}\frac{n^n}{n!} x^{n}=\frac{-W(-x)}{1+W(-x)} $$ $$ \sum_{n\geq 1}\frac{n^n}{n!} x^{2n+1}=\frac{-x W(-x^2)}{1+W(-x^2)} $$ $$ \sum_{n\geq 1}\frac{n^n}{n!}(2n+1) x^{2n}=\frac{-x W(-x^2)}{1+W(-x^2)} $$ $$ \sum_{n\geq 1}\frac{n^n}{n!}(2n+1) x^{n}=\frac{-3W(-x)-2W(-x)^2-W(-x)^3}{(1+W(-x))^3} $$ and by replacing $x$ with $x(1-x)$, then applying $\int_{0}^{1}(\ldots)\,dx$, we get $$\begin{eqnarray*} \sum_{n\geq 1}\frac{n^n n!}{(2n)!} &=& \int_{0}^{1}\frac{-3W(-x(1-x))-2W(-x(1-x))^2-W(-x(1-x))^3}{(1+W(-x(1-x)))^3}\,dx\\ &=&\int_{0}^{1}\frac{-3W\left(\frac{t^2-1}{4}\right)-2W\left(\frac{t^2-1}{4}\right)^2-W\left(\frac{t^2-1}{4}\right)^3}{\left(1+W\left(\frac{t^2-1}{4}\right)\right)^3}\,dt\\&=&2\int_{-1/4}^{0}\frac{-3W(t)-2W(t)^2-W(t)^3}{\sqrt{1+4t}(1+W(t))^3}\,dt\\&=&\int_{W(-1/4)}^{0}-\frac{2 e^x x (3+2x+x^2)}{(1+x)^2 \sqrt{1+4 e^x x}}\,dx\end{eqnarray*}$$ where the last function is clearly integrable on $\left(W(-1/4),0\right)$. This integral representation is also a good way for deriving that the value of the wanted series is $\approx 1.53261$. An alternative derivation of an upper bound comes from the Cauchy-Schwarz inequality and the identities $$ \sum_{n\geq 1}\frac{n!^2}{(2n)!}e^n = \frac{e}{4-e}+\frac{4\sqrt{e}}{(4-e)^{3/2}}\arcsin\left(\frac{\sqrt{e}}{2}\right),$$ $$ \sum_{n\geq 1}\frac{n^{2n}}{(2n)!e^n}=\frac{1}{2}\left(\frac{1}{1+W\left(\frac{1}{2\sqrt{e}}\right)}\right).$$