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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!

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1 Answer 1

up vote 1 down vote accepted

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? –  Joe Berg Feb 26 '12 at 2:09
    
@Gingerjin: I don't understand your question. Can you please elaborate? –  Aryabhata Feb 26 '12 at 2:11
    
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. –  Joe Berg Feb 26 '12 at 2:14
    
@Gingerjin: Yes it is possible, I think. After cancelling one $n$, it looks like a Riemann Sum. –  Aryabhata Feb 26 '12 at 2:19
2  
@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 $a_n=\log\left(\frac{3n-1}{3n+3}\right)\to0$. –  Did Feb 26 '12 at 11:24

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