Question

How do you test the alternating series `Sigma (-1)^n/rootn n` from n is `[2,oo)` for convergence?

Answer 1

The alternating series:

`sum_(n=2)^oo (-1)^n/root(n)(n)`

is not convergent.

Explanation:

This is an alternating series so we can apply Leibniz's test stating that the series is convergent if given:

`a_n = 1/(root(n)(n)`

we have:

(i) `lim_(n->oo) a_n = 0`

(ii) `a_(n+1)/a_n < 1`

Now consider the sequence:

`b_n = ln a_n = ln (1/(root(n)(n))) = -1/n ln n`

We have:

`lim_(n->oo) b_n = 0`

And:

`lim_(n->oo) a_n =lim_(n->oo) e^(b_n)`

but as `e^x` is a continuous function:

`lim_(n->oo) a_n = e^(lim_(n->oo) b_n) = e^0 = 1`

The Leibniz's test is then not satisfied and the series is not convergent.