Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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

2 Answers 2

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

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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