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I'm taking a class on Sequences and Series and there's a question in my homework that goes like this:

Given the definition:

A sequence $(a_n)$ is said to converge to $L$ if and only if $\forall \epsilon > 0$, there exists $N \in \mathbb{N}$ such that $n > N \Rightarrow |a_n - L| < \epsilon$.

Prove that the sequence $(\frac{7n+5}{3n-5})$ converges to $\frac{7}{3}$. There's a hint that asks to solve the inequality $3n-5 > n$ first, and I've done so, but I'm not sure how it helps me.

I have tried breaking down the equation $|(\frac{7n+5}{3n-5})- \frac{7}{3}| < \epsilon $ and got kind of stuck in a circular algebra loop. I've also tried plugging in $n = 5/2$ (the solution to $3n-5 > n$ ) in order to get something meaningful out of it but I'm not seeing anything.

I don't want the answer to the question, but just a hint as to what to do next. I know that the original inequality has something to do with the solution to this, but I just can't figure out what...

Thanks!

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When you compute $\frac{7n+5}{3n-5} - \frac73$, you get an expression $\frac{a}{b(3n-5)}$. The inequality $3n-5 > n$ serves for you to be able to estimate that fraction by the simpler fraction $\frac{a}{bn}$. –  Daniel Fischer Sep 18 '13 at 19:52

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up vote 3 down vote accepted

When we combine the fractions, we get $$\frac{7n+5}{3n-5}-\frac73=\frac{21n+15}{9n-15}-\frac{21n-35}{9n-15}=\frac{50}{9n-15}=\frac{50}{3(3n-5)}.$$ Since $3n-5>n$ when $n>2,$ then $$\left|\frac{7n+5}{3n-5}-\frac73\right|<\frac{50}{3n}$$ for $n>2.$ Can you take it from there?

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Yes, I can see it now! Thank you! –  Benjamin Kovach Sep 18 '13 at 20:02

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