# Product of infinite terms $\lim_{n\to\infty}\left(\frac{(n+1)(n+2)…(3n)}{n^{2n}}\right)$

Let there be a series

$$\lim_{n\rightarrow \infty}\left(\frac{(n+1)(n+2)...(3n)}{n^{2n}}\right)$$

is equal to?

For this type of problem I am unable to approach.The numerator starts from $(n+1)$ to $3n$ where as the denominator has ${n^{2n}}$ terms

• Thank for editing, this question is correct – Samar Imam Zaidi Sep 18 '17 at 3:14

One possibility that I usually do for products is to take logarithms. For any finite $n$, we have $$\ln\left(\frac{(n+1)\dots(3n)}{n^{2n}}\right)=\sum_{k=1}^{2n}\ln\left(\frac{n+k}{n}\right)=\sum_{n=1}^{2n}\ln\left(1+\frac{k}{n}\right)=n\sum_{k=1}^{2n}\frac{\ln(1+\tfrac{k}{n})}{n}$$ In the limit, the sum $$\sum_{k=1}^{2n}\frac{\ln(1+\tfrac{k}{n})}{n}$$ approaches a positive, nonzero definite integral, whereas $n$ approaches infinity. Thus the logarithm approaches infinity, so the original product does as well.

• Can be converted into Integral form using the property of Definite Integral – Samar Imam Zaidi Sep 18 '17 at 3:55
• I want to confifm with you whether the answer is $\int_{0}^{2} log(1+x)dx$ – Samar Imam Zaidi Sep 18 '17 at 4:15
• @SamarImamZaidi Yes, that is the correct integral – TomGrubb Sep 18 '17 at 4:29

There are $2n$ factors in the numerator. Of them, the first $n$ are each greater than $n$, and the next $n$ are each greater than $2n$. Thus, the numerator is greater than $n^n (2n)^n$. The denominator is $n^{2n}$ and so the ratio is greater than $2^n$, which of course tends to $\infty$ as $n\to\infty$.

• $\infty$ for $n\to\infty$ is not the answer, i presume that we can arrive by using log and definite integral concept – Samar Imam Zaidi Sep 18 '17 at 3:29
• That's what you think. – kimchi lover Sep 18 '17 at 3:34
• @SamarImamZaidi As you can see from other answers, the limit indeed is $+\infty$. Comparing with the answers posted so far, I consider this approach the most elegant, +1 from me. – Martin Sleziak Sep 18 '17 at 5:14

Hint:) With Stirling's approximation $$n!\sim\sqrt{2\pi n}\left(\dfrac{n}{e}\right)^n$$ we have $$\lim_{n\rightarrow \infty}\left(\frac{(n+1)(n+2)...3n}{n^{2n}}.\dfrac{n!}{n!} \right)=\lim_{n\rightarrow \infty}\dfrac{(3n)!}{n!n^{2n}}=\lim_{n\rightarrow \infty}\sqrt{3}\left(\dfrac{27}{e^2}\right)^n=\color{blue}{\infty}$$

Let us denote $$a_n=\frac{(n+1)(n+2)...(3n)}{n^{2n}}.$$

Then you have $$\frac{a_{n+1}}{a_n} = \frac{(3n+1)(3n+2)(3n+3)}{(n+1)^3} \left(\frac{n}{n+1}\right)^{2n}.$$

I should be relatively easy to check that $$\lim\limits_{n\to\infty} \frac{a_{n+1}}{a_n} > 1.$$ (In fact, the limit is equal to $27/e^2$ unless I missed something.)

Can you tell something about $a_n$ combining the facts that it is positive and starting from some $n_0$ you have $a_{n+1}/a_n>1+\varepsilon$ (for some $\varepsilon>0$)?

In the other words, can you show this: If $a_n>0$ for each and $\lim\limits_{n\to\infty} a_{n+1}/a_n>1$, then $\lim\limits_{n\to\infty} a_n=\infty$?

• Very elegant solution without using Stirling. +1 – Paramanand Singh Sep 18 '17 at 8:26