If $(a_n)$ is decreasing and $a_{n+1}+a_n\to0$ then $a_n\to0$ 
Suppose that $(a_n)$ is a decreasing sequence of numbers such that $\displaystyle{\lim_{n\rightarrow +\infty} (a_{n+1}+a_n)=0}$. 
  I want to show that $\displaystyle{\lim_{n\rightarrow +\infty} a_n=0}$. 

I have done the following: 
Since the sequence $(a_n)$ is decreasing, we have that $$a_{n+1}\leq a_n \Rightarrow a_{n+1}-a_n\leq 0$$ 
Then 
$$\begin{align}
\lim_{n\rightarrow +\infty}(a_{n+1}-a_n)\leq 0 &\Rightarrow \lim_{n\rightarrow +\infty}(a_{n+1}+a_n-a_n-a_n)\leq 0 \\
&\Rightarrow \lim_{n\rightarrow +\infty}(a_{n+1}+a_n-2a_n)\leq 0 \\
&\Rightarrow \lim_{n\rightarrow +\infty}(a_{n+1}+a_n)-2\lim_{n\rightarrow +\infty} a_n\leq 0 \\
&\Rightarrow -2\lim_{n\rightarrow +\infty} a_n\leq 0
\end{align}$$ 
Is this correct so far? 
How could we continue? 
Or is there an other way to get the desired result? 
 A: Hint:
Since $a_n$ is decreasing you have
$$   a_{n+1}+a_n \leq a_n +a_n \leq a_n +a_{n-1}$$
A: By dividing the last inequality by $-2$, we clearly get the following:
$$\lim_{n \to +\infty} a_n \geq 0$$
To prove that this is equal to $0$, we now only need to prove that:
$$\lim_{n \to +\infty} a_n \leq 0$$
Now, if we shift the sequence to be $a_{n+1}$ rather than $a_n$, then we have not changed what value it converges to as $n \to +\infty$ because we just shifted the sequence. Thus, we have the following:
$$\lim_{n \to +\infty} a_{n+1}=\lim_{n \to +\infty} a_n$$
Thus, we will prove instead that the left hand side of this equation is less than or equal to $0$, which will prove the whole theorem. Now, to do this, we will show that $\lim_{n \to +\infty} a_{n+1} \leq \lim_{n \to +\infty} a_n+a_{n+1}$ or something of the like. To do this, we need to somehow relate $a_{n+1}$ to $a_n+a_{n+1}$ in an inequality. Let's try breaking up $a_{n+1}$ into two parts:
$$a_{n+1}=\frac 1 2(a_{n+1}+a_{n+1})$$
Now, as you said, $a_n \geq a_{n+1}$. Thus, if we substitute the second $a_{n+1}$ with $a_n$ in the right hand side, we get an expression greater than or equal to the left-hand side, meaning, as @Did said.
$$a_{n+1} \leq \frac 1 2(a_{n+1}+a_n)$$
Thus, the limit as these two expressions as $n \to +\infty$ have the same relation:
$$\lim_{n \to +\infty} a_{n+1} \leq \lim_{n \to +\infty} \frac 1 2(a_{n+1}+a_n)$$
Bring the $\frac 1 2$ out of the limit on the right-hand side:
$$\lim_{n \to +\infty} a_{n+1} \leq \frac 1 2\lim_{n \to +\infty} a_{n+1}+a_n$$
It is given that $\lim_{n \to +\infty} a_{n+1}+a_n=0$, so the right-hand side is just $0$:
$$\lim_{n \to +\infty} a_{n+1} \leq 0$$
Thus, by showing that this limit is less than or equal to $0$ and using the inequality that we derived from your work at the top, we have:
$$\lim_{n \to +\infty} a_n = 0$$
