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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.

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  • $\begingroup$ Your expression for $\log f(n)$ is incorrect, should be $$\log(1/n)+(1/n)\log(\dots)$$ $\endgroup$ – Gerry Myerson Aug 25 '13 at 13:14
  • $\begingroup$ Which is $f(n)$ supposed to be? $$ \frac1n\left\{\frac{(2n+1)(2n+2)}{(2n+n)}\right\}^{1/n} $$ or $$ \frac1n\left\{(2n+1)(2n+2)-(2n+n)\right\}^{1/n} $$ $\endgroup$ – robjohn Aug 25 '13 at 13:29
  • $\begingroup$ Oh, I see, it is an extended product. $\endgroup$ – robjohn Aug 25 '13 at 13:35
  • $\begingroup$ latex anyone please ? $\endgroup$ – Arjang Aug 25 '13 at 13:38
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    $\begingroup$ @Arjang: I was already doing it, thus my question :-) $\endgroup$ – robjohn Aug 25 '13 at 13:43
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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 $$\int_0^1\ln(x+2)dx=3\ln3-1-\{2\ln2-0\}=\ln (3^3)-\ln e-\ln(2^2)=\ln \frac{27}{4e}$$

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  • $\begingroup$ What about the power '1/n' of the whole series in the question? $\endgroup$ – Rajath Krishna R Aug 25 '13 at 15:13
  • $\begingroup$ @RajathKrishnaR, as $\ln (a^m)=m\ln a$ $\endgroup$ – lab bhattacharjee Aug 25 '13 at 15:39
  • $\begingroup$ Sir, but in the question itself there is one 1/n.......so taking log and applying above rule would make it 1/n^2...... $\endgroup$ – Rajath Krishna R Aug 25 '13 at 15:53
  • $\begingroup$ @RajathKrishnaR, as there are $n$ terms as the multipliers, $$\displaystyle f(n)=\frac1n\Big\{(2n+1)(2n+2)\cdots(2n+n)\Big\}^{1/n}=\left(\prod_{1\le r\le n}\frac{2n+r}n\right)^{\frac1n}$$ Now apply log wrt $e$ $\endgroup$ – lab bhattacharjee Aug 25 '13 at 15:58
  • $\begingroup$ Ohhh.....now I get it sir....thank you for your answer..... $\endgroup$ – Rajath Krishna R Aug 25 '13 at 16:06
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Hint (for this multiple choice question): $2 \le f(n) \le 3$.

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  • $\begingroup$ Hah, multiple choice does make this problem easier, I guess. :) $\endgroup$ – Thomas Andrews Aug 26 '13 at 13:53
  • $\begingroup$ njguliyev how you get $2 \leq f(n)\leq 3,$ explain me , thanks $\endgroup$ – DXT Jan 19 '17 at 10:23
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$$\frac{1}{n}\left[\frac{(3n)!}{(2n)!}\right]^{\frac{1}{n}}\sim\frac{1}{n}\left[\frac{\left(\frac{3n}{e}\right)^{3n}\sqrt{6\pi n}}{\left(\frac{2n}{e}\right)^{2n}\sqrt{4\pi n}}\right]^{\frac{1}{n}}=\frac{1}{n}\left[\left(\frac{n}{e}\right)^n\frac{27^n}{4^n}\sqrt{\frac{3}{2}}\right]^{\frac{1}{n}}\to \frac{27}{4e}$$

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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:

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