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The limit in question $$\lim_{n\to \infty} \frac{n}{n!!^\frac{2}{n}}.$$

What I have done: $$ \lim_{n\to \infty}\frac{(n^\frac{2}{n})^\frac{n}{2}}{n!!^\frac{2}{n}}. $$

Then taking the logarithm and converting it into a summation series

$$\lim_{n\to \infty}\frac{2}{n}\sum_{r=0}^{n-1}\ln\frac{1}{1-\frac{2r}{n}}.$$

Substituting $\frac{r}{n} =t$ and converting it into an integral yields

$$-2\int_{0}^1 \ln(1-2t)dt.$$

Integrating gives this function

$$-2[(t+1)\ln(1-2t) -t]_{0}^1.$$

Can someone help me find the limit?

calculatormathematical
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1 Answers1

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If $n=2k$ then $n!!=(2k)!!=2^kk!$.

Thus, $$\dfrac{n}{(n!!)^{2/n}}=\dfrac{2k}{2(k!)^{1/k}}=\dfrac{k}{(k!)^{1/k}}$$

But one can show (for example, see this) that $$\lim_{k\to\infty} \dfrac{k}{(k!)^{1/k}}=e $$

If $n=2k-1$ then $n!!=(2k-1)!!=\dfrac{(2k-1)!}{2^{k-1}(k-1)!}$ and so $$\dfrac{n}{(n!!)^{2/n}}=\dfrac{(2k-1)2^{\frac{2(k-1)}{(2k-1)}}[(k-1)!]^{2/(2k-1)}}{[(2k-1)!]^{2/(2k-1)}}$$

But the limit of this expression must be the same as that of $$2\cdot\dfrac{(2k-1)}{[(2k-1)!]^{1/(2k-1)}}\cdot \dfrac{[(k-1)!]^{2/(2k-1)}}{[(2k-1)!]^{1/(2k-1)}}$$

The first factor goes to $e$ by the above limit. The second has the same limit as $$\dfrac{[(k-1)!]^{1/(k-1)}}{[(2k-1)!]^{1/(2k-1)}}$$

And again, by the above this has the same limit as $$\dfrac{\dfrac{k-1}{e}}{\dfrac{2k-1}{e}}\longrightarrow \dfrac{1}{2} $$

So for $n=2k-1$ the sequence converges to $e$ too.

As both the odd and even subsequences converge to $e$, we conclude $$\dfrac{n}{(n!!)^{2/n}}\to e$$

Julio Puerta
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