# Calculate $\lim_{n\rightarrow\infty}\frac{1}{n}\left(\prod_{k=1}^{n}\left(n+3k-1\right)\right)^{\frac{1}{n}}$

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!

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

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

Therefore,

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

• 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? – user475680 Apr 30 '14 at 9:50
• @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. – Ron Gordon Apr 30 '14 at 9:53