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Per the title of this question, how does one go about calculating $$\lim_{n\rightarrow \infty} \frac{1}{n^2}\left(\ln\left(\frac{2^n}{3^n}\right)+\ln\left(\frac{5^n}{4^n}\right)+\cdots+\ln\left(\frac{(3n-1)^n}{(n+2)^n}\right)\right)\ ?$$

Thanks!

1 Answers 1

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For a mechanical* way to do this:

Hint:

$$\log \frac{(3k-1)}{k+2} = \log 3 + \log (1- \frac{1}{3k}) - \log (1 + \frac{2}{k})$$

and

$$\log (1-x) = -x - \frac{x^2}{2} - \frac{x^3}{3} - \dots $$

for $|x| \lt 1$.

More details:

Using the above hint, the $k^{th}$ term is $$ n \log \frac{(3k-1)}{k+2} = n \log 3 + \frac{5n}{3k} + \mathcal{O}(\frac{n}{k^2})$$

and so your sum is

$$ \frac{1}{n^2} \sum_{k=1}^{n} (n \log 3 + \frac{5n}{3k} + \mathcal{O}(\frac{n}{k^2}))$$

$$ = \frac{1}{n^2}(n^2\log 3 + \frac{5n\log n}{3} + \mathcal{O}(n))$$

Here we used the estimate $\displaystyle \sum_{k=1}^{n} \frac{1}{k} = \log n + \mathcal{O}(1)$ and $\displaystyle \sum_{k=1}^{n} \frac{1}{k^2} = \mathcal{O}(1)$

Thus the sum is $$ = \log 3 + \frac{5\log n}{3n} + \mathcal{O}(\frac{1}{n})$$

and so your limit is $$\log 3$$

Note that we don't really need the estimate $\displaystyle \sum_{k=1}^{n} \frac{1}{k} = \log n + \mathcal{O}(1)$

All we need to show is that $\displaystyle \sum_{k=1}^{n} \frac{1}{k} = o(n)$ and this easily follows from the following classic theorem:

If $\displaystyle a_n \to 0$, then $\displaystyle \frac{1}{n} \sum_{k=1}^{n} a_k \to 0$

*As Didier points out (see comments below), this last theorem can be used to skip all the mechanical calculations done above by applying it directly to the terms of your sequence.

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    If I consider the sequence as the integer, can I work it out?2012-02-26
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    @Gingerjin: I don't understand your question. Can you please elaborate?2012-02-26
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    I mean I want to create an integral that is the answer of the sequence, Can I work it out? SOrry, I haven't explained it correctly.2012-02-26
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    @Gingerjin: Yes it is possible, I think. After cancelling one $n$, it looks like a Riemann Sum.2012-02-26
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    Could you please show me that? I couldn't work it out.3ks2012-02-26
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    @Gingerjin: Maybe I was wrong. We need to get $\frac{k}{n}$ in there somehow.2012-02-26
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    Thanks. Is it true that Taylor expansion is commonly used in this kind of problems?2012-02-26
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    @Gingerjin: Well, it is a useful and powerful tool. If you are talking about in exams or something like that, the answer is, it depends on the teacher I guess.2012-02-26
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    @Aryabhata: Could you add in more detail? I can't quite see the entire solution yet (in case this is a concern - this isn't homework, just a question from a practice exam ;). Thanks!2012-02-26
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    @josh: Perhaps if you edit the question with your attempt at completing this answer, I can help you better. Knowing where exactly you are having problems will help us focus on the right things. Otherwise I will have to try and cover all possibilities... btw, if you hover your mouse over the block at the end of the answer, you will see more detail.2012-02-26
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    @Aryabhata: For the most part I don't understand the second step in the 'complete answer' section, where you go on to claim the complete sum without $n^2$ in the denominator is (such and such). Why does the $log(n)$ appear there, and why do we get the $O(n)$? (I'm used to remainder in the form of $R_4(n)$ moreso than big-O or small-O)2012-02-26
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    @josh: I am using $\sum_{k=1}^{n} \frac{1}{k} = \log n + \gamma + O(1/n)$ and the fact that $\sum_{k=1}^{\infty} \frac{1}{k^2}$ converges and so $\sum_{k=1}^{n} \frac{1}{k^2} = O(1)$.2012-02-26
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    Ah, okay. Thank you. We have not learned the 1+1/2+...->ln(n) approximation in this course. Or well, we have, since we learned about Taylor polynomials, but I guess I just don't understand the big-O notation quite well.2012-02-26
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    @josh: You don't really need to know that approximation to solve this problem. All you need to prove is that $\sum_{k=1}^{n} \frac{1}{k} = o(n)$ (note: small Oh, not big Oh). For that: This is a well-known theorem: if $a_n \to 0$, then $\frac{1}{n}\sum_{k=1}^{n} a_k \to 0$.2012-02-26
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    @Aryabhata Your $k$ in the answer should be an $n$. When you sum from $n\,\log 3$, the index of summation is $n$, and the sum $n(n-1)\log3/2$. The limit is $\log3/2$, as can also be seen using [Stolz's criterium](http://en.wikipedia.org/wiki/Stolz%E2%80%93Ces%C3%A0ro_theorem).2012-02-26
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    @JuliánAguirre: If you see the first version, that is what I had. But I realized that the question is actually $\lim_{n \to \infty} \frac{1}{n^2} \sum_{k=1}^{n} n \log (\frac{3k-1}{k+2})$ and not $\lim_{n \to \infty} \frac{1}{n^2} \sum_{k=1}^{n} k \log (\frac{3k-1}{k+2})$ which gives $\frac{\log 3}{2}$2012-02-26
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    @Aryabhata I see.2012-02-26
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    @Aryabatha: the argument in the very last line of your post allows to skip everything before since the sum one is interested in is $(\log3)+\frac1n\sum\limits_{k=1}^na_k$ with $a_n=\log\left(\frac{3n-1}{3n+3}\right)\to0$.2012-02-26
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    @DidierPiau: Yes agreed.2012-02-26