Is there any result that could help me obtain a lower bound for this series? The series has the structure
$\sum_{k=1}^\infty p_ka_k$
and converges to some $c<0$.
$\{a_k\}_{k=1}^\infty$ is known, is an increasing function of $k$ with $a<a_k<0, \forall k$, has limit $0$ and known sum $\sum_{k=1}^\infty a_k=A$.
$\{p_k\}_{k=1}^\infty$, on the other hand, is unknown apart from the fact that $p_k \in [0,1],\forall k$ and $\sum_{k=1}^\infty p_k = 1$.
I'm interested in whether it is possible to obtain a lower bound for $c$, possibly depending on the form of $\{p_k\}_{k=1}^\infty$.
Any advice?
This is my first post here, hope I didn't break any rule!
Thanks in advance for you patience!
EDIT
My actual question could be also posed as 'can I prove,given what's above, that $p_1=1,p_k=0, k>1$ is the sequence that yields the greatest lower bound for $c$, which would then equal $a_1$?'
 A: I might have come up with something, I'm asking for your opinion on this.
Given the working hypotheses, $\sum_k p_k a_k$ can be seen as the expected value of $a_k$ w.r.t. the probability law $p$ of $k$. If I recall correctly, Cauchy internality should hold, by which
$E_p[a_k] \in [\inf(a_k),\sup(a_k)]$
thus leading to $\inf(a_k)=a_1$, since $a_k$ is a strictly increasing function of $k$ on $k=1,2,...$, and this should hold whatever the $p_k$s... 
Moreover, the sequence of $p_k$ that yields exactly that lower bound is the one that concentrates mass 1 on $k=1$, i.e. $p_1=1, p_k=0, k>1$, so this should be (one of?) the 'worst possible scenario'..
What do you guys think?
Thanks again!
A: I will attempt a proof that $a_1$ is the greatest lower bound for $c$ and that it is generated uniquely by $p_1=1, \forall n(p_n=0)$.

Theorem 1. $a_1$ is a lower bound.
Proof. Because $(a_n)_{n=1}^\infty$ is an increasing sequence,
$$\sum_{k=1}^\infty p_ka_k\ge \sum_{k=1}^\infty p_ka_1= a_1$$

Theorem 2. $a_1+h$ is not a lower bound, for any $h>0$.
Proof. Assume $a_1+h$ is a lower bound. Then, for any sequence $(p_n)_{n=1}^\infty$ with $\forall n\in\Bbb N_1(p_n\in[0,1])$ and $\sum_{n=1}^\infty p_n=1$,
$$\sum_{k=1}^\infty p_ka_k\ge a_1+h$$
Choose the sequence such that $p_1=1$ and $p_n=0$ for all $n>1$. Then it directly follows that $a_1\ge a_1+h$, which is a contradiction.

Theorem 3. The only $(p_n)_{n=1}^\infty$ for which $c=a_n$ is $p_1=1,\forall n(p_n=0)$. In other words, if $p_1\ne1$, then there exists an $h>0$ such that $a_1+h$ is a lower bound.
Proof. Let $p_1\ne1$. Then, pick any $n>1$ such that $p_n>0$ (of which there must be at least $1$). Now let $h=p_n(a_n-a_1)>0$. It then follows that
$$\sum_{k=1}^\infty p_ka_k=\sum_{k=1}^{n-1} p_ka_k+p_na_n+\sum_{k=n+1}^\infty p_ka_k=\sum_{k=1}^{n-1} p_ka_k+p_na_1+h+\sum_{k=n+1}^\infty p_ka_k\\
\ge\sum_{k=1}^{n-1} p_ka_1+p_na_1+h+\sum_{k=n+1}^\infty p_ka_1=\sum_{k=1}^\infty p_ka_1+h=a_1+h$$

Let me know if you find any mistakes or have any questions. I think it works.
