Showing $\lim_{n \to \infty}\;( (2n+1)! )^{1/n}\,\sin\left(\frac {1}{n^2-n}\right) = \frac{4}{e^2}$

Question

$ \displaystyle \lim_{n \to \infty}( (2n+1)! )^{1/n} \sin(\frac {1}{n^2-n}) $= $\frac{4}{e^2}$ the factorial part tends to $ \infty$ while the sin part to $0$

considering that sin is continous : $ \lim_{n\to\infty} \sin(\frac {1}{n^2-n})=0$ because the denominator tends to infinity ( the $n^2$ grows asimptotically faster than $n$ )

while using the quotion theorem we can find $ a_n:= (( 2n+1)!)^{1/n}$ let $ \lambda = \frac{a_{n+1}}{a_n}= \frac{(( 2n+3)!)^{1/(n+1)}}{(( 2n+1)!)^{1/n}} > \frac{(( 2n+3)!)^{1/(n+1)}}{(( 2n+1)!)^{1/n+1}} = ( ( (2n+3)(2n+2 ) ) ^{1/n+1} = (4n^2 +10n +6)^{n+1}=(1+(4n^2 +10n +5 ))^{ \frac{1}{4n^2 +10n +5} (4n +10 +5/n)} = e^\infty = \infty$

therefore $ \lambda \rightarrow \infty>0 \implies a_n \rightarrow \infty$

now i know that the limit is of the form $ \infty 0$

Answer 1

Hint:

You can use Stirling's formula: $$m!\sim m^me^{-m}\sqrt{2\pi m}$$ and the small angle approximation for the sine function: $$\sin \left(\frac1{n^2-n}\right)\sim \frac1{n^2-n}$$

Answer 2

You can use the following theorem:

Theorem: Given any sequence $\,(y_{n})_{n\geqslant1}\,$ of (strictly) positive real numbers, suppose that $\lim\limits_{n\to\infty}{\dfrac{y_{n+1}}{y_{n}}}$ exists (finite or infinite), then

$\lim\limits_{n\to\infty}{\sqrt[{n}]{y_{n}}}=\lim\limits_{n\to\infty}{\dfrac{y_{n+1}}{y_{n}}}.$

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For any $\;n\in\Bbb N\setminus\{1\}\;,\;$ it results that

$\big((2n+1)!\big)^{1/n}\sin\left(\dfrac1{n^2-n}\right)=$

$=\sqrt[n]{(2n+1)!}\cdot\!\dfrac{\sin\left(\frac1{n^2-n}\right)}{\frac1{n^2-n}}\!\cdot\!\dfrac{n^2}{n^2-n}\!\cdot\!\dfrac1{n^2}=$

$=\sqrt[n]{\dfrac{(2n+1)!}{n^{2n }}}\cdot\!\dfrac{\sin\left(\frac1{n^2-n}\right)}{\frac1{n^2-n}}\!\cdot\!\dfrac{n^2}{n^2-n}\;.$

Now we will consider the sequence $\;y_n=\dfrac{(2n+1)!}{n^{2n }}\;.$

$\lim\limits_{n\to\infty}\dfrac{y_{n+1}}{y_n}=\lim\limits_{n\to\infty}\dfrac{(2n+3)!\,n^{2n}}{(n+1)^{2n+2}\,(2n+1)!}=$

$=\lim\limits_{n\to\infty}\dfrac{(2n+2)(2n+3)}{(n+1)^2\left(1+\frac1n\right)^{2n}}=\dfrac4{e^2}\;.$

Consequently, by applying the previous theorem, we get that there exists the limit :

$\lim\limits_{n\to\infty}\sqrt[n]{y_n}=\lim\limits_{n\to\infty}\sqrt[n]{\dfrac{(2n+1)!}{n^{2n}}}=\dfrac4{e^2}\;.$

Moreover,

$\lim\limits_{n\to\infty} \left[\big((2n+1)!\big)^{1/n}\sin\left(\dfrac1{n^2-n}\right)\right]=$

$=\lim\limits_{n\to\infty}\left[\sqrt[n]{\dfrac{(2n+1)!}{n^{2n }}}\cdot\!\dfrac{\sin\left(\frac1{n^2-n}\right)}{\frac1{n^2-n}}\!\cdot\!\dfrac{n^2}{n^2-n}\right]=$

$=\dfrac4{e^2}\cdot1\cdot1=\dfrac4{e^2}\;.$

Answer 3

The idea of the OP needs a little tweak. The sequence in the limit cab be rewritten as $$\begin{align} ( (2n+1)! )^{1/n} \sin(\frac {1}{n^2-n})=\frac{\sqrt[n]{(2n+1)!}}{(n^2-n)}\frac{\sin\big((n^2-n)^{-1}}{(n^2-n)^{-1}} \end{align} $$

Since $(n^2-n)^{-1}\rightarrow0$ as $n\rightarrow\infty$, $$\lim_{n\rightarrow\infty}\frac{\sin\big((n^2-n)^{-1}}{(n^2-n)^{-1}}=1$$

To estimate the limit $\lim_n\frac{\sqrt[n]{(2n+1)!}}{(n^2-n)}$ recall a well known result from Calculus: $$\liminf_n\frac{a_{n+1}}{a_n}\leq \liminf_n\sqrt[n]{a_n}\leq\limsup_n\sqrt[n]{a_n}\leq\limsup_n\frac{a_{n+1}}{a_n}$$ This idea is already expressed by the poster the question. Letting $$\begin{align} a_n=\frac{(2n+1)!}{(n^2-n)^n} \end{align} $$ we have $$ \frac{a_{n+1}}{a_n}=\frac{(2n+3)(2n+2)}{(n+1)n}\Big(1-\frac{2}{n+1}\Big)^n $$ Now, $\lim_{n\rightarrow\infty}\Big(1-\frac{2}{n+1}\Big)^n=e^{-2}$, and $\lim_{n\rightarrow\infty}\frac{(2n+3)(2n+2)}{n(n+1)}=4$ All this shows that the limit in the OP exists and that its value is $4e^{-2}$

Answer 4

As an alternative, starting as already suggested, we have

$$( (2n+1)! )^{1/n} \sin\left(\frac {1}{n^2-n}\right)=\frac{\sqrt[n]{(2n+1)!}}{(n^2-n)}\frac{\sin\big((n^2-n)^{-1}}{(n^2-n)^{-1}} $$

with $\frac{\sin\big((n^2-n)^{-1}}{(n^2-n)^{-1}} \to 1$ and

$$\frac{\sqrt[n]{(2n+1)!}}{(n^2-n)}=e^{\frac{\log((2n+1)!)-n\log(n^2-n)}{(n+1)-n}} \to \frac4{e^2}$$

which can be obtained by Stolz-Cesaro theorem indeed

$$\frac{\log((2(n+1)+1)!)-(n+1)\log((n+1)^2-(n+1))-\log((2n+1)!)+n\log(n^2-n)}{n}=$$

$$=\log\left(\frac{(2n+3)(2n+2)}{n^2+n}\right)-n\log\left(\frac{n^2+n}{n^2-n}\right)=$$

$$=\log\left(\frac{4n^2+10n+6}{n^2+n}\right)-\log\left[\left(\left(1+\frac{2n}{n^2-n}\right)^{\frac{n^2-n}{2n}}\right)^{\frac{2n^2}{n^2-n}}\right]\to \log 4-\log(e^2) = \log 4 -2$$

which implies

$$\frac{\log((2n+1)!)-n\log(n^2-n)}{n}\to \log 4 -2$$