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$$\lim_{n\to\infty} \left({\frac{2n+1}{3n+2}}\right)^{n} = ? $$

Simple method gives an indeterminate expression. Any idea how to think out this case?

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4 Answers 4

up vote 1 down vote accepted

Probably a bit cumbersome but straightforward:

$$ \lim_{n \to \infty} \bigg(\frac{2n +1}{3n+2} \bigg)^n=\lim_{n \to \infty}\bigg( \frac{2}{3} \cdot \bigg(\frac{n+\frac{1}{2}}{n+\frac{2}{3}} \bigg)\bigg)^n = \lim_{n \to \infty}\bigg( \frac{2}{3} \cdot \bigg(\frac{n+\frac{2}{3}-\frac{1}{6}}{n+\frac{2}{3}}\bigg)\bigg)^n =\lim_{n \to \infty}\bigg( \frac{2}{3} \cdot \bigg(1-\frac{1}{6n+4} \bigg)\bigg)^n= \lim_{n \to \infty}\bigg( \frac{2}{3} \bigg)^n \lim_{n \to \infty} \bigg(1-\frac{1}{6n+4} \bigg)^{(6n +4-4)\cdot \frac{1}{6}}=0 \cdot e^{-\frac{1}{6}} \cdot 1 = 0 $$

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HINT: $$0<\frac{2n+1}{3n+2}<\frac23$$ for $n\ge 1$.

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3  
+1 So painfully simple –  DonAntonio Dec 5 '12 at 1:11
    
Probably sounds quit cheesy, but officially your solution is much better than mine) –  Alex Dec 7 '12 at 0:02
    
@Alex: Apparently not for the OP! Yours does have a certain baroque impressiveness. :-) –  Brian M. Scott Dec 7 '12 at 0:06

Brian's answer is right on the money. Another way to see it is noting the algebraic equality

$$\left(\frac{2n+1}{3n+2}\right)^n=e^{n\log\frac{2n+1}{3n+2}}$$

But

$$n\log\frac{2n+1}{3n+2}\xrightarrow [n\to\infty]{}-\infty$$

since $\,\log\frac{2}{3}<0\,$ , so the limit's like

$$\lim_{m\to-\infty}e^m=0$$

But I'd choose Brian's answer as it is way simpler...:)

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if f(n) is less then a constant k which is less then 1, for large enough n, and all proceeding n, then as n tends to infinity $f(n)^n$=0, so the limit of your expression is zero.

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1  
Maybe could change to less than $k$ which is less than $1$, to make sure the criterion is not used on, for example, $(1-1/n)^n$. –  André Nicolas Dec 5 '12 at 1:04

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