Proving a series converges given that it is bounded below If $\sum{p_n}$ is a convergent series such that $p_i \in \mathbb{R}$ $\forall i$ and {$q_n$} is a sequence that is bounded below where $q_i \in \mathbb{R}$ $\forall i$, I want to show that if $$q_{n+1} \leq q_n+p_n$$ then {$q_n$} converges. 
What I know:
Given that {$q_n$} is bounded below, if I can show that it is monotonic then by definition it converges. 
Also I know that if $\sum{p_n}$ converges then $\lim_{n \to \infty}p_n = 0$ 
I know I'm missing a key step to show that $q_{n+1} \leq q_n$, if anyone could help I would be very grateful!
 A: Note $q_1+\dots+q_n = q_1+(q_2-q_1)+\dots+(q_n-q_{n-1}) \le q_1+p_1+\dots+p_n$ and so $\sum q_n \le q_1+ \sum p_n$ and thus converges since $q_n$ is bounded from below.
Also, $(q_n)_n$ does not have to be monotonic. If $p_n$ is positive, then $q_{n+1}$ can be in between $q_n$ and $q_n+p_n$ or less than $q_n$.
A: 
The following proof is a generalization of that given by @zhw. in This Solution.


Given Information:
The sequence $q_n$ satisfies the conditions that $(i)$ 
$$q_{n+1}-q_n\le p_n\tag 1$$
for some sequence $p_n$ such that $\sum_{k=1}^\infty p_k$ converges, and $(ii)$ $q_n$ is bounded below.

Implication of Convergence of $\,\,\,\,\,\displaystyle \sum_{n=1}^\infty p_n$
Since $\sum_{n=1}^\infty p_n$ converges, then from the Cauchy criterion, for any $\epsilon>0$, there exists a number $N$ such that whenever $N''>N'>N$, $$\sum_{n=N'}^{N''}p_n<\epsilon/2\tag 2$$

Consequence of Lower Bound on $q_N$
Since $q_n$ is bounded below, then its limit inferior exists and is finite.  We denote the limit inferior as $q$ and write $\displaystyle q=\liminf_{n\to \infty}q_n$.  
Then, for all $\epsilon>0$, there exists a number $N$ such that 
$$q_n>q-\epsilon\tag 3$$
for all $n>N$.  
Furthermore, for all $\epsilon>0$, there exists a subsequence $n_k$ such that 
$$ \displaystyle q_{n_k}<q+\epsilon/2 \tag4$$

Putting It Together:
Now, given $\epsilon>0$, we see from $(2)$ that there exists a $k_0$ large enough so that $\sum_{n_{k_0}}^{n_{k_0}+m}p_n <\epsilon/2$.
And we have from $(1)$
$$\begin{align}
\sum_{n=n_{k_0}}^{n_{k_0}+m}(q_{n+1}-q_n)&=q_{n_{k_0}+m}-q_{n_{k_0}}\\\\
&\le \sum_{n=n_{k_0}}^{n_{k_0}+m}p_n\\\\
&<\epsilon/2
\end{align}$$
whereupon rearranging yields
$$q_{n_{k_0}+m}-q\le (q_{n_{k_0}}-q)+\epsilon/2 \tag 5$$
Applying $(4)$ to $(5)$ shows that
$$q_{n_{k_0}+m}-q\le \epsilon \tag 6$$
Finally, combining $(3)$ and $(6)$, we find that given $\epsilon>0$, 
$$-\epsilon< q_n-q < \epsilon$$
whenever $n>N=n_{k_0}$.  

And this completes the proof that $q_n$ converges!

A: This is a solution from one of my friends.
Define $c_n=q_n-q_{n+1}$, proving $\sum_{n=1}^\infty |c_n|<\infty$ will gives us that $\{q_n\}$ is a Cauchy sequence, and thus is convergent. For this reason, it is sufficent to prove $\sum_{n=1}^\infty |c_n|<\infty$.
It should be notice that $-c_n\le p_n$ since $q_{n+1}\le q_n+p_n$. Define $N_1=\{n\in\mathbb{N}|c_n\ge 0\}$ and $N_2=\{n\in\mathbb{N}|c_n< 0\}$. Then for each $n\in\mathbb{N}$, we have
\begin{align*}
\sum_{k=1}^n|c_k|&=\sum_{k\in N_1,k\le n}c_k-\sum_{k\in N_2,k\le n}c_k\\
&=\sum_{k=1}^nc_k-2\sum_{k\in N_2,k\le n}c_k\\
&\le q_1-q_{n+1}+2\sum_{k\in N_2,k\le n}p_k\\
&\le q_1-q+2P,
\end{align*}
where $q$ is a lower bound of $\{q_n\}$ and $P=\sum_{k=1}^\infty p_k<\infty$. Since this holds for all $n$, we have $\sum_{n=1}^\infty |c_n|<\infty$, the proof is complete.
