Prove that $\displaystyle \frac{1}{1}-\frac{1}{2^2}+\frac{1}{3}-\frac{1}{4^2}+...$ diverges.
that's alternative series .... use Leibniz test
$\displaystyle \displaystyle \sum _{n=1} ^{\infty} (-1)^n a_n $
if :
$\displaystyle \displaystyle \lim_{n\to \infty} a_n = 0 $
and
$\displaystyle a_n \ge a_{n+1} $
than it converges if not than diverges
(or show that converge/diverge by the definition)
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 \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 \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 \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. $
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 \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 \displaystyle \sum _{n=0} ^{\infty} \frac {1}{2n+1} - \frac {1}{(2n+2)^2 } $
diverge