## Series: test Leibnitz

Investigate series to which it is possible to apply the test of Leibnitz and investigate the convergence of this method. To investigate which of them is possible to investigate the convergence to TEST OF ABSOLUTE CONVERGENCE

1) $\sum_{n=0}^{\infty} (-1)^n \frac{1}{2^n}$

My solution:

$b_{n+1} = \frac{1}{2^{n+1}} < b_n$
$\lim_{n \to \infty} \frac{1}{2^n} = 0$
Is Convergent

Test of absolute convergence:
$|\sum_{n=0}^{\infty} (-1)^n \frac{1}{2^n}|$
$\frac{1}{2^n}$ is convergent, I prove in the retio test:
$\lim_{n \to \infty} \frac{\frac{1}{2^{n+1}}}{\frac{1}{2^n}} = \frac{1}{2} < 0$

2) $\sum_{n=0}^{\infty} (-1)^n \frac{(n+1)}{3^n}$

$\lim_{n \to \infty} \frac{n+1}{3^n} = \infty$
Is divergence

Series divergence, test of absolute convergence can not be applied

3) $\sum_{n=0}^{\infty} (-1)^n \frac{1}{\sqrt{n+1}}$

$b_{n+1} < b_n$
$\lim_{n \to \infty} = 0$
Is convergence

Test of absolute convergence:
I use test of integral:
$\int_1^{\infty} \frac{1}{\sqrt{x+1}}dx = \infty$

Test of absolute convergence can not be applied

The problems are correct ?