# Using the definition of convergence, how do you prove that the sequence (-1)^n/(n^3-ln(n)) converges from n=1 to infinity?

Nov 10, 2016

Take $N > {\left(\frac{2}{\epsilon}\right)}^{\frac{1}{3}}$

#### Explanation:

The limit is 0 so

$\left\mid {s}_{n} - L \right\mid = \frac{1}{{n}^{3} - \ln n} < \frac{1}{{n}^{3} - n} = \frac{1}{n \left({n}^{2} - 1\right)} < \frac{1}{n \cdot {n}^{2} / 2} = \frac{2}{n} ^ 3$

Now for each $n > N > {\left(\frac{2}{\epsilon}\right)}^{\frac{1}{3}}$ we get

$\left\mid {s}_{n} - L \right\mid < \frac{2}{n} ^ 3 < \frac{2}{{\left(\frac{2}{\epsilon}\right)}^{\frac{1}{3}}} ^ 3 = \epsilon$