How to compute $\lim_{n\rightarrow\infty}\frac1n\left\{(2n+1)(2n+2)\cdots(2n+n)\right\}^{1/n}$

Question

If $\displaystyle f(n)=\frac1n\Big\{(2n+1)(2n+2)\cdots(2n+n)\Big\}^{1/n}$, then $\lim\limits_{n\to\infty}f(n)$ equals: $$ \begin{array}{} (\mathrm{A})\ \frac4e\qquad&(\mathrm{B})\ \frac{27}{4e}\qquad&(\mathrm{C})\ \frac{27e}{4}\qquad&(\mathrm{D})\ 4e \end{array} $$ I couldn't get the right way to start off with this problem. But, as the options include the constant $e$ I think I will have to work out with logarithms.

So, this is what I did. $$ \lim_{n \to \infty} f(n)\\ =\mathrm{exp}\left(\ln\left(\lim_{n \to \infty} f(n)\right)\right)\\ =\mathrm{exp}\left(\lim_{n \to \infty}\left[\ln\left(f(b)\right)\right]\right)\\ =\mathrm{exp}\left(\lim_{n \to \infty}\left[\frac{1}{n^2}\left(\ln(2n+1)+\ln(2n+2)+\ldots+\ln(2n+n)\right)\right]\right) $$ I'm not able to proceed further. In case my method is correct please give me hints on proceeding further and in case it is wrong give me the same on another method.

Answer 1

As there are $n$ terms as the multipliers,

$$\displaystyle f(n)=\frac1n\Big\{(2n+1)(2n+2)\cdots(2n+n)\Big\}^{1/n}=\left(\prod_{1\le r\le n}\frac{2n+r}n\right)^{\frac1n}$$

hence $$\ln f(n)=\frac1n\sum_{1\le r\le n}\ln\left(2+\frac rn\right)$$ Using $$\lim_{n \to \infty} \frac1n\sum_{r=1}^n g\left(\frac rn\right)=\int_0^1g(x)dx,$$

$$\lim_{n\to\infty}\ln f(n)=\int_0^1\ln(x+2)dx$$

Note that \begin{eqnarray} \int\ln(x+2)dx&=&x\ln(x+2)-\int\frac x{x+2}dx\\ &=& x\ln(x+2)-\int\frac{x+2-2}{x+2}dx\\ &=& x\ln(x+2)-\int\ dx+2\int\frac1{x+2}dx\\ &=& x\ln(x+2)-x+2\ln(x+2)\\ &=&(x+2)\ln(x+2)-x \end{eqnarray} hence $$\int_0^1\ln(x+2)dx=3\ln3-1-\{2\ln2-0\}=\ln (3^3)-\ln e-\ln(2^2)=\ln \frac{27}{4e}$$

Answer 2

Hint (for this multiple choice question): $2 \le f(n) \le 3$.

Answer 3

$$\frac{1}{n}\left[\frac{(3n)!}{(2n)!}\right]^{\frac{1}{n}}\sim\frac{1}{n}\left[\frac{\left(\frac{3n}{e}\right)^{3n}\sqrt{6\pi n}}{\left(\frac{2n}{e}\right)^{2n}\sqrt{4\pi n}}\right]^{\frac{1}{n}}=\frac{1}{n}\left[\left(\frac{n}{e}\right)^n\frac{27^n}{4^n}\sqrt{\frac{3}{2}}\right]^{\frac{1}{n}}\to \frac{27}{4e}$$

Answer 4

We can express $f(n)=\sqrt[n]{a_n}$ for $$a_n=\frac{(2n+1)(2n+2)\dots(2n+n)}{n^n}.$$ Let us check whether the sequence $a_{n+1}/a_n$ has a limit. \begin{align*} \frac{a_{n+1}}{a_n} &= \frac{(3n+3)(3n+2)(3n+1)}{(2n+1)(2n+2)} \cdot \frac{n^n}{(n+1)^{n+1}}\\ \frac{a_{n+1}}{a_n} &= \frac{(3n+3)(3n+2)(3n+1)}{(2n+1)(2n+2)(n+1)} \cdot \frac{n^n}{(n+1)^n}\\ \frac{a_{n+1}}{a_n} &= \frac{(3n+3)(3n+2)(3n+1)}{(2n+1)(2n+2)(n+1)} \cdot \frac1{\left(1+\frac1n\right)^n} \end{align*} From this we see that $$\lim\limits_{n\to\infty} \frac{a_{n+1}}{a_n} = \frac{27}{4e}.$$

Now if this limit exists, then the limit $\lim\limits_{n\to\infty} \sqrt[n]{a_n}$ also exists and has the same value. See:

• Limit of ${a_n}^{1/n}$ is equal to $\lim_{n\to\infty} a_{n+1}/a_n$
• Inequality involving $\limsup$ and $\liminf$: $ \liminf(a_{n+1}/a_n) \le \liminf((a_n)^{(1/n)}) \le \limsup((a_n)^{(1/n)}) \le \limsup(a_{n+1}/a_n)$