I'm need of some assistance regarding a homework question:

"calculate the following: $\lim_{n\rightarrow\infty}\frac{1}{n}\left(\prod_{k=1}^{n}\left(n+3k-1\right)\right)^{\frac{1}{n}}$"

Alright so since this question is in the chapter for definite integrals (and because it is similar to other questions I have answered) I assumed that I should play a little with the expression inside the limit and change the product to some Riemann sum of a known function.

OK so I've tried that but with no major breakthroughs...

Any hints and help is appreciated, thanks!

  • 1
    $\begingroup$ Have you tried logarithms? $\endgroup$ – 5xum Apr 30 '14 at 9:02
  • $\begingroup$ Yes and I wasn't able to come up with something useful $\endgroup$ – user475680 Apr 30 '14 at 9:31

The product $P_n$ may be expressed as follows:

$$P_n = \left [ \prod_{k=1}^n \left (1+\frac{3 k-1}{n}\right ) \right ]^{1/n} $$

so that

$$\log{P_n} = \frac1{n} \sum_{k=1}^n \log{\left (1+\frac{3 k-1}{n}\right )}$$

as $n \to \infty$, $P_n \to P$ and we have

$$\log{P} = \lim_{n \to \infty} \frac1{n} \sum_{k=1}^n \log{\left (1+\frac{3 k-1}{n}\right )} = \lim_{n \to \infty} \frac1{n} \sum_{k=1}^n \log{\left (1+\frac{3 k}{n}\right )}$$

which is a Riemann sum for the integral

$$\log{P} = \int_0^1 dx \, \log{(1+3 x)} = \frac13 \int_1^4 du \, \log{u} = \frac13 [u \log{u}-u]_1^4 = \frac{8}{3} \log{2}-1$$


$$P = \frac{2^{8/3}}{e} $$

  • $\begingroup$ Ah I see now, really cool solution, but I don't understand this line: $\lim_{n \to \infty} \frac1{n} \sum_{k=1}^n \log{\left (1+\frac{3 k-1}{n}\right )} = \lim_{n \to \infty} \frac1{n} \sum_{k=1}^n \log{\left (1+\frac{3 k}{n}\right )}$. How do you get rid of the (-1) in the nominator? $\endgroup$ – user475680 Apr 30 '14 at 9:50
  • 1
    $\begingroup$ @user475680: because the log is continuous and you may bring the limit operator inside the log, so you essentially are taking the limit of $1/n$, which obviously vanishes. $\endgroup$ – Ron Gordon Apr 30 '14 at 9:53

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