A question about convergence for series and sequence of real numbers The exercise is stated below: 

Let $\sum_{}^{} b_n $ be a series of real numbers which converges, and let
  {$a_n$} be a sequence of real numbers which is bounded below. Prove
  that $a_{n+1} \leq a_n + b_n$ $\implies$ {$a_n$} converges.

I want to show that {$a_n$} is a cauchy sequence then the conclusion will follow. So I did $a_{n+1} - a_n \leq b_n$. Since $\sum_{}^{} b_n $ converges, for large enough N, $n \geq N \implies b_n  = 0$. Then take this N and let $n \geq N $ so that $a_{n+1} - a_n$ is inifnitely small. Therefore, it is a cauchy sequence. 
But I thought my proof is not rigorous at all since I could not specify the exact $\epsilon$ or N that I need to show a cauchy sequence. Could you please help me with this proof?
 A: There is a mistake in your proof:
You showed that 
$$a_{n+1} - a_n \leq b_n,$$
but we do not have 
$$0 \leq a_{n+1} - a_n$$
and we cannot deduce that $\{a_n\}$ is Cauchy sequence.
But if we define
$$c_n=a_n+\sum_{k\geq n}b_k,$$
then from assumption we must have
$$c_{n+1}\leq c_n.$$
Since $\{a_n\}$ is bounded below we have $\{c_n\}$ is bounded below and thus converges. Then from the convergence of $\{c_n\}$ we deduce the the convergence of $\{a_n\}.$
A: You've correctly admitted that your proof is not rigorous. Indeed, $\sum b_n$ converges does not imply the existence of such an $N$.
Instead, use the given recurrence to say
\begin{align*}
a_{n+1} \leq a_n + b_n \leq a_{n-1} + b_{n-1} + b_n \leq \dots \leq \sum_1^n b_k
\end{align*}
which converges. Thus the set of partial sums $\{\sum_1^n b_k\}$ is bounded above by some constant (if not, we could find a sequence of partial sums that tends to infinity, a contradiction). This means $\{a_k\}$ is also bounded above by the inequalities shown above. 
Given any subsequence of $\{a_k\}$, by Bolzano Weierstrauss, we can find a convergent further subsequence (since $\{a_k\}$ is bounded both above and below). 
Since any subsequence of $\{a_k\}$ has a further subsequence which converges, the entire sequence must converge.
