Convergence Tests Hi I'm doing these questions:
Determine whether the following series converges and if it does, find its sum
i) $\sum_{k=1}^{\infty}\frac{1}{9k^2+3k-2}$
ii)$\sum_{k=1}^{\infty}\ln \left( \frac{k+2}{k+1} \right)$
I just need a hint on where to start. What test do I use? 
For i) I started by doing Comparison test. So I'm comparing it to $\frac{1}{k^2}$ then I realized it's a p series and concluded that because the power of k > 1, the series converges therefore $\sum_{k=1}^{\infty}\frac{1}{9k^2+3k-2}$ also converges. Is this even right? And how do I find its sum?
 A: I have resolved first one i) :
$$
\sum_{k=1}^{\infty} { 1 \over {9k^2+3k-2}}={1 \over 3} \sum_{k=1}^{\infty}{({1 \over {3k-1}} - {1 \over {3k+2}})}={1 \over 3} \sum_{k=1}^{\infty}{({1 \over {3k-1}} - {1 \over {3(k+1)-1}})}={1 \over 3} (({1 \over 2} - {1 \over 5}) + ({1 \over 5}-{1 \over 8})+({1 \over 8}-{1 \over 11})+...)={1 \over 6}
$$
The second one is infinite :
$$
\sum_{k=1}^\infty ln({{k+2}\over{k+1}})=\sum_{k=1}^\infty ({ln(k+2)-ln(k+1)})=(ln(3)-ln(2))+(ln(4)-ln(3))+(ln(5)-ln(4))+(ln(6)-ln(5))+...= \lim_{k \to \infty} \ln(k) - \ln(2) = \infty
$$
I hope it help you. Daniel
A: For a start,
looking at
$9k^2+3k-2
$,
the discriminant is
$3^2-4(9)(-2)
=9+72
=81
$.
Since this is a square,
you can factor it.
So its roots are
$\dfrac{-3\pm 9}{18}
=\dfrac{-1\pm 3}{6}
=-\frac23, \frac13
$,
from which you get the factorization
(as Daniel Pol uses)
$9k^2+3k-2
=(3k+2)(3k-1)
$.
Looking at Daniel Pol's
partial fraction decomposition,
I noticed that
it was of the form
$\dfrac1{ak-b}-\dfrac1{ak+a-b}
=\dfrac{(ak+a-b)-(ak-b)}{(ak-b)(ak+a-b)}
=\dfrac{a}{a^2k^2+ak(a-2b)-b(a-b)}
$
where
$0 < b < a$.
Therefore
$\dfrac{1}{a^2k^2+ak(a-2b)-b(a-b)}
=\dfrac1{a}\left(\dfrac1{ak-b}-\dfrac1{ak+a-b}\right)
$.
Yours is the case
$a=3, b=1$.
This produces a
telescoping sum
since
$\begin{array}\\
\sum_{k=1}^n\left(\dfrac1{ak-b}-\dfrac1{ak+a-b}\right)
&=\sum_{k=1}^n\left(\dfrac1{ak-b}-\dfrac1{a(k+1)-b}\right)\\
&=\sum_{k=1}^n\dfrac1{ak-b}-\sum_{k=1}^n\dfrac1{a(k+1)-b}\\
&=\sum_{k=1}^n\dfrac1{ak-b}-\sum_{k=2}^{n+1}\dfrac1{ak-b}\\
&=\dfrac1{a-b}-\dfrac1{a(n+1)-b}\\
&\to\dfrac1{a-b}
\qquad\text{as } n \to \infty\\
\end{array}
$
so that
$\sum_{k=1}^{\infty} \dfrac{1}{a^2k^2+ak(a-2b)-b(a-b)}
=\dfrac1{a(a-b)}
$.
