Does Liebniz-Criteria become a necessary condition for convergence if $a_n$ is monotonically decreasing? The Leibniz Criterion says that if the sequence $a_n$ is monotonically decreasing then the following statements are equivalent:

\begin{align}
1) &amp; &amp; &amp; \sum_{n=0}^\infty (-1)^na_n \text{ converges} \\[6pt]
2) &amp; &amp; &amp; \lim_{n\rightarrow\infty}(a_n) = 0
\end{align}

so this has to mean that if $a_n$ is monotonically decreasing, the following is true

$$  \sum_{n=0}^\infty (-1)^na_n \text{ converges} \Leftrightarrow \lim_{n\rightarrow\infty}(a_n) = 0 $$

I know that $ \lim_{n\rightarrow\infty}(a_n) = 0 $ is not a necessary condition for convergence if $a_n$ is not monotonically decreasing, but the way I am reading the rule stated in my script, it becomes a necessary condition if $a_n$ is monotonically decreasing.

For example to prove the non-convergence of 

$$ \sum_{n=0}^\infty (-1)^n(1)_n $$

I can just point out to the fact that 

$$ \lim_{n\rightarrow\infty}(1)_n = 1$$

because $(1)_n$ is a monotonically decreasing sequence.
Is this interpretation correct?
As Giuseppe Negro has noticed, for the convergence of $\displaystyle \sum a_n$,$$ \lim_{n\rightarrow\infty}(a_n) = 0 $$ is a necessary condition anyway.
The convergence test of Leibniz is only applicable to alternating series, i.e. series where every positive term is followed by a negative term, and every negative term is followed by a positve term ( + - + - + - + - ...).

For a general series, the condition $\displaystyle\lim_{n \to \infty} a_n= 0$ is a necessary condition for convergence, but not sufficient. This is the main point of Leibniz test. He could prove that for alternating series this condition is both necessary and sufficient. Thus if you've got an alternating series $\sum (-1)^n a_n$ with $\displaystyle\lim_{n \to \infty} a_n= 0$, than you are certain it converges. 

I don't know why you are referring to the condition of "monotonically decreasing". This doesn't matter for Leibniz test. It's only an issue in the integral test.