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

Jan 26, 2017

The series is convergent.

#### Explanation:

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

(i) ${\lim}_{n \to \infty} {a}_{n} = 0$

(ii) ${a}_{n + 1} < {a}_{n}$ for $n > N$

then the series is convergent.

In our case:

${\lim}_{n \to \infty} \frac{{2}^{n} + 1}{{3}^{n} - 2} = {\lim}_{n \to \infty} \frac{{\left(\frac{2}{3}\right)}^{n} - \frac{1}{3} ^ n}{1 - \frac{2}{3} ^ n} = 0$

so the first condition is satisfied.

Now consider:

${a}_{n + 1} = \left(\frac{{2}^{n + 1} + 1}{{3}^{n + 1} - 2}\right) = \frac{2}{3} \left(\frac{{2}^{n} + \frac{1}{2}}{{3}^{n} - \frac{2}{3}}\right)$

clearly:

$\left({2}^{n} + \frac{1}{2}\right) < \left({2}^{n} + 1\right)$

and:

$\left({3}^{n} - \frac{2}{3}\right) > \left({3}^{n} - 2\right)$

so that:

$\left(\frac{{2}^{n} + \frac{1}{2}}{{3}^{n} - \frac{2}{3}}\right) < \left(\frac{{2}^{n} + 1}{{3}^{n} - 2}\right)$

and it follows that:

${a}_{n + 1} = \left(\frac{{2}^{n + 1} + 1}{{3}^{n + 1} - 2}\right) = \frac{2}{3} \left(\frac{{2}^{n} + \frac{1}{2}}{{3}^{n} - \frac{2}{3}}\right) < \frac{2}{3} \left(\frac{{2}^{n} + 1}{{3}^{n} - 2}\right) < {a}_{n}$

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