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We have the chain $f: \mathbb{N} \rightarrow \mathbb{R}$ where $f(n)= \frac{2}{7}, \frac{5}{12} \cdots \frac{3n-1}{5n+2}$ and $f(1) =0$, so I have to find the limit and study the convergence.

If the function is convergent,that means it has a limit. That limit in my opinion is 3/5?

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The second one :) – dfsfds Feb 5 '13 at 21:39
There is a sequence $f(n)=(3n-1)/(5n+2)$ and $f(1)=2/7\neq 0$ the limit is $3/5$ – Adi Dani Feb 5 '13 at 21:41
Shouldnt just we take the highest power in the denominator and numerator? and it is 3/5? – dfsfds Feb 5 '13 at 21:42
Just dividing top and bottom by $n$ is sufficient to see that the limit exists and has value $3/5$. – David Mitra Feb 5 '13 at 21:42

how rigorous is the answer supposed to? The simple proof is that $$ \lim_{n \to \infty} \frac{3n -1}{5n +2} = \lim_{n \to \infty} \frac{3-\frac{1}{n}}{5+\frac{2}{n}}=\frac{3}{5} $$

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Though a simple question, students get confused because they are institutionalized.It happens and check-point for student.

Looking at the last term of the series,numerator: 3 times a " extreme larger quantity" Denominator : 5 times a "extreme large quantity". The " extreme larger quantity" cancels out leaving 3/5.

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