Using the Partial Summation Formula Partial Summation formula:
Consider $\sum a_n$ and $\sum b_n$.  If $A_n= \sum _{k=1}^{n} a_k$, then  
$\sum _{k=1}^{n} a_kb_k = A_nb_{n+1}- \sum _{k=1}^{n} A_k(b_{k+1}-b_k)$  
So $\sum _{k=1}^{\infty} a_kb_k$ converges if both
$\sum _{k=1}^{\infty} A_k(b_{k+1}-b_k)$ 
and $\{A_nb_{n+1}\}$ converge
The problem I'm working on is:
Given that $\sum c_n$ converges where each $c_n > 0$ prove that
$\sum (c_nc_{n+1})^{1/2}$ also converges.  
I wanted to use the partial summation formula to help me solve this.
I let $\{a_n\}=(c_n)^{1/2}$ and $\{b_n\}=(c_{n+1})^{1/2}$
Since $\sum c_n$ converges, $\lim_{n\to\infty} a_n = 0$ which implies $\lim_{n\to\infty} b_n = 0$
hence we get $\{A_nb_{n+1}\}$ converges
I'm getting stuck at proving $\sum _{k=1}^{\infty} A_k(b_{k+1}-b_k)$ converges.
The second part of the problem says:
Show that the converse is also true if $\{c_n\}$ is monotonic.
I'm not really sure where to start on this one, but if $\{c_n\}$ is monotonic and already bounded below by $0$ it has to be decreasing in order for $\sum (c_nc_{n+1})^{1/2}$ to converge.

Off the top of my head, partial summation is not what I would use to solve your first problem.  Instead note that for all $n$
$\sqrt{c_n c_{n+1}} \leq \sqrt{ (\max c_n, c_{n+1})^2} = \max c_n, c_{n+1} \leq c_n + c_{n+1}$.  
By a direct comparison, this implies $\sum_n \sqrt{ c_n c_{n+1}}$ converges.
Is there some reason you are trying to use partial summation?  Were you instructed to?
As for the second part, note that if $\{c_n\}$ is decreasing, then $c_n c_{n+1} \geq c_{n+1}^2$.  This should help to make a comparison going "the other way"...
