# Thread: Struggling to prove convergene using definition of convergence

1. ## Struggling to prove convergene using definition of convergence

Hi, first post, found myself stuttering a bit so just trying to get my head round some examples of proving certain sequences converge.

The first one, I think I did correctly, prove that

(an) = (5n^2+srt(n))/n^2 converges to five

Obviously you have to show that the modulus of an-l is less than episilon for all n.
I found myself rearranging (an)-l and got (an-l)=n^-(3/2).
So then I took epsilon to be an arbitary real number, greater than sqrt(1/N^3)

The second one, I'm struggling to even put it into a suitable format for an-l, it is show that the sequence:

(an)= (6+5n)/(2-3n) converges to -5/3.
It's probably pretty basic algebra, but I'm struggling to even get this into the form where I can pick an arbitrary epsilon and take an N.
So could do with some help on this one.

Apologies if the typing looks a bit ungainly, I'll try my best to fit in here

2. Originally Posted by MickHarford
The first one, I think I did correctly, prove that

(an) = (5n^2+srt(n))/n^2 converges to five
Obviously you have to show that the modulus of an-l is less than episilon for all n.
I found myself rearranging (an)-l and got (an-l)=n^-(3/2).
So then I took epsilon to be an arbitary real number, greater than sqrt(1/N^3)
Actually we start with [b]any[/u] $\varepsilon > 0$.
We can see that $\dfrac{1}{{\sqrt[3]{{\varepsilon ^2 }}}} > 0$ so
$\left( {\exists N} \right)\left[ {n \geqslant N\, \Rightarrow \,n \geqslant \dfrac{1}{{\sqrt[3]{{\varepsilon ^2 }}}}} \right]$.
Then we see $\dfrac{1}{{\sqrt {n^3 } }} < \varepsilon$