Question

How do you test the alternating series `Sigma (-1)^n(2^n+1)/(3^n-2)` from n is `[0,oo)` for convergence?

Answer 1

The series is convergent.

Explanation:

A sufficient condition for an alternating series to converge is established by the Leibniz test stating that if:

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

(ii) `a_(n+1) < a_n` for `n > N`

then the series is convergent.

In our case:

`lim_(n->oo) (2^n+1)/(3^n-2) = lim_(n->oo) ((2/3)^n -1/3^n)/(1-2/3^n) = 0`

so the first condition is satisfied.

Now consider:

`a_(n+1) = ((2^(n+1)+1)/(3^(n+1)-2)) = 2/3 ((2^n+1/2)/(3^n-2/3))`

clearly:

`(2^n+1/2) < (2^n+1)`

and:

`(3^n-2/3) > (3^n-2)`

so that:

`((2^n+1/2)/(3^n-2/3)) <((2^n+1)/(3^n-2))`

and it follows that:

`a_(n+1) = ((2^(n+1)+1)/(3^(n+1)-2)) = 2/3 ((2^n+1/2)/(3^n-2/3)) < 2/3((2^n+1)/(3^n-2)) < a_n`

Thus also the second condition is satisfied and the series os convergent.