did i prooved this convergense corretly..

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an = h − 1h2\displaystyle a_n\, =\, \frac{h\, -\, 1}{h^2}

lim⁡h→+∞ an = lim⁡h→+∞ h − 1h2 = lim⁡h→+∞ hh2 − 1h2h2h2 =0 − 01 = 0\displaystyle \lim_{h\rightarrow +\infty}\, a_n\, =\, \lim_{h\rightarrow +\infty}\, \frac{h\, -\, 1}{h^2}\, =\, \lim_{h\rightarrow +\infty}\, \frac{\frac{h}{h^2}\, -\, \frac{1}{h^2}}{\frac{h^2}{h^2}}\, = \frac{0\, -\, 0}{1}\, = \, 0

∣an − L∣ < ϵ\displaystyle |a_n\, -\, L|\, <\, \epsilon

∣h − 1h2 − 0∣ < ϵ\displaystyle \left|\frac{h\,-\,1}{h^2}\,-\,0\right|\,<\,\epsilon

∣h − 1h2∣ = h − 1h2 = 1h − 1h2 < 1h < ϵ   ⟺   1ϵ < h\displaystyle \left|\frac{h\,-\,1}{h^2}\right|\,=\,\frac{h\,-\,1}{h^2}\,=\,\frac{1}{h}\,-\,\frac{1}{h^2}\,<\,\frac{1}{h}\,<\,\epsilon\,\iff\,\frac{1}{\epsilon}\,<\,h

N = [1ϵ] + 1\displaystyle N\, =\, \left[\frac{1}{\epsilon}\right]\, +\, 1

N > 1ϵ\displaystyle N\, >\, \frac{1}{\epsilon}
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Think that h might be an n, ;).

It looks okay to me... except you did the proof backwards. You are to pull an N "out of thin air" when you present a proof.

So let e>0. Let N be the unique positive integer such that N >= 1/e > (N-1). Then if n > N ....
 
transgalactic said:
in the end what do i need to write precisely regarding the N part?
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All proofs like this start:
"Let epsilon > 0."
"Let N = f(epsilon)."
"Assume n > N."

The "f(epsilon)" should be the maximum of any assumptions on N (if any) and the expression containing epsilon you solved for in your scrap work.

You could say exactly what I wrote, or something like "Let N be the smallest (or any) integer havinging the property that N >= 1/epsilon." Since epsilon is a fixed positive real number and the integers are not bounded above, we may assume such an N exists.
 
so at the start i assume e>0
then i write what i wrote in y first post

what is N = f(epsilon) in this case ??

i can say N>1/e
 
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