1
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ Thank for editing, this question is correct $\endgroup$ – Samar Imam Zaidi Sep 18 '17 at 3:14
3
$\begingroup$

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.

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

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 $n\to\infty$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ $\infty$ for $n\to\infty$ is not the answer, i presume that we can arrive by using log and definite integral concept $\endgroup$ – Samar Imam Zaidi Sep 18 '17 at 3:29
  • $\begingroup$ That's what you think. $\endgroup$ – kimchi lover Sep 18 '17 at 3:34
  • $\begingroup$ @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. $\endgroup$ – Martin Sleziak Sep 18 '17 at 5:14
3
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

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 $\lim\limits_{n\to\infty} a_n=\infty$?

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Very elegant solution without using Stirling. +1 $\endgroup$ – Paramanand Singh Sep 18 '17 at 8:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.