1. ## Prove divergence

Prove that $\frac{1}{1}-\frac{1}{2^2}+\frac{1}{3}-\frac{1}{4^2}+...$ diverges.

2. Originally Posted by alexmahone
Prove that $\frac{1}{1}-\frac{1}{2^2}+\frac{1}{3}-\frac{1}{4^2}+...$ diverges.
that's alternative series .... use Leibniz test

$\displaystyle \sum _{n=1} ^{\infty} (-1)^n a_n$

if :

$\displaystyle \lim_{n\to \infty} a_n = 0$

and

$a_n \ge a_{n+1}$

than it converges if not than diverges

(or show that converge/diverge by the definition)

3. Originally Posted by yeKciM
that's alternative series .... use Leibniz test

$\displaystyle \sum _{n=1} ^{\infty} (-1)^n a_n$

if :

$\displaystyle \lim_{n\to \infty} a_n = 0$

and

$a_n \ge a_{n+1}$

than it converges if not than diverges

(or show that converge/diverge by the definition)
The alternating series test can be used to prove convergence, not divergence.

4. Originally Posted by alexmahone
The alternating series test can be used to prove convergence, not divergence.
if series do not converge than, series diverge .... just must state that "based on Leibniz test series diverge" (or show by the definition, using partial sums )

5. Originally Posted by yeKciM
if series do not converge than, series diverge .... just must state that "based on Leibniz test series diverge" (or show by the definition, using partial sums )
If the alternating series test works, the series converges. But if the alternating series test fails, we cannot say anything about the convergence or divergence of the series.

6. Originally Posted by alexmahone
If the alternating series test works, the series converges. But if the alternating series test fails, we cannot say anything about the convergence or divergence of the series.
hm... when some series converge means that there is all members of the series except them finite many are in region of some point... if that's not the case than it diverges ... but ... never mind that I'm probably off while translating

try like this :

$\displaystyle \sum _{n=0} ^{\infty} \frac {1}{2n+1} - \frac {1}{(2n+2)^2 }$

and show that diverges by definition of using some of the test's for convergence (Cauchy, D'Alambert , Raabe or by the integral test )

D'Alambert

$\displaystyle q= \lim_{n \to \infty } \frac {a_{n+1}}{a_n} = \left\{\begin{matrix}
q<1 & converges \\
q>1 & diverges \\
q=1 & ???
\end{matrix}\right.$

Raabe :

$\displaystyle q= \lim_{n \to \infty } n(\frac {a_{n}}{a_{n+1} }-1 ) = \left\{\begin{matrix}
q>1 & converges \\
q<1 & diverges \\
q=1 & ???
\end{matrix}\right.$

7. Liebniz's test does not apply here because $a_n\ge a_{n+1}$ is not true.

8. Originally Posted by HallsofIvy
Liebniz's test does not apply here because $a_n\ge a_{n+1}$ is not true.
yes, my bad ... didn't wrote it like (actually didn't see the series correctly at that time) in the post #6

Liebniz's test could not be used because this is not the alternative series

$\displaystyle \sum _{n=0} ^{\infty} (-1)^n a_n$

Cauchy and D'alambert's test give value of one so it's inconclusive but using Raabe's or integral test series

$\displaystyle \sum _{n=0} ^{\infty} \frac {1}{2n+1} - \frac {1}{(2n+2)^2 }$

diverge