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

1 Answer
Jun 18, 2018

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.