Question

Using the definition of convergence, how do you prove that the sequence `lim 1/(6n^2+1)=0` converges?

Answer 1

Given any number `epsilon > 0` choose `M > 1/sqrt(6epsilon)`, with `M in NN`.

Then, for `n >= M` we have:

`6n^2+1 > 6n^2 > 6M^2 >= 6/(6epsilon) = 1/epsilon`

and so:

`n>= M=> 1/(6n^2+1) < epsilon`

which proves the limit.