# 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) & & & \sum_{n=0}^\infty (-1)^na_n \text{ converges} \\[6pt] 2) & & & \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?

• The condition $b_n\to 0$ is always necessary to the convergence of $\sum b_n$. When $b_n=(-1)^n a_n$, one has the equivalence $b_n\to 0\ \iff\ a_n\to 0$. So $a_n\to 0$ is a necessary condition for the convergence of $\sum (-1)^n a_n$. May 31 '15 at 11:49
• Yes the statement is true. Proof: $$b_n\to 0\ \iff\ \lvert b_n\rvert \to 0 \iff \lvert (-1)^n a_n\rvert\to 0 \iff a_n\to 0.$$ If you are not convinced please post a counterexample. May 31 '15 at 11:59
• if it were so why would leibniz go through the trouble of putting the additional monotonic decreasing condition there May 31 '15 at 11:59
• I guess you are mixing up "necessary" and "sufficient" here. $a_n\to 0$ is necessary to the convergence of $\sum (-1)^na_n$. $a_n\to 0$ AND $a_n$ monotonically decreasing is sufficient to the convergence of $\sum (-1)^n a_n$. May 31 '15 at 12:01
• I know realize that you were right May 31 '15 at 12:33

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.

• I understand your first point, but on your second point$\sum (-1)^n a_n$ with $\displaystyle\lim_{n \to \infty} a_n= 0$ you are certain that is converges, only when a is monotonically decreasing. The rule is defined for monotonically decreasing a May 31 '15 at 12:36
• @Steven Van Geluwe: Alternation of signs and terms going to $0$ is not sufficient for convergence. Example, let $a_n=\frac{1}{n}$ if $n$ is odd, and let $a_n=\frac{1}{2^n}$ when $n$ is even. Then $\sum_1^\infty (-1)^n a_n$ diverges. May 31 '15 at 12:52
• @Gravity: There are many series $\sum (-1)^n a_n$ that converge even though the $a_n$ are not monotonically decreasing. For example, let $a_n=\frac{1}{n^2}$ when $n$ is odd, and let $a_n=\frac{1}{n^3}$ when $n$ is even. May 31 '15 at 12:59
• @Andre: that is correct and that is why Leibniz says the condition is sufficient but not necessary, but still we can only apply the Leibniz rule if $a_n$ is monotonically decreasing. we cannot in all Circumstances say $\sum (-1)^n a_n$ converges when $\displaystyle\lim_{n \to \infty} a_n= 0$ is true, and that is the point May 31 '15 at 13:05
• I down voted, because I am afraid that the main point of this answer is wrong. Leibniz's test requires monotonicity of the $a_n$ term. May 31 '15 at 21:25