Let us consider an alternating series $~~\displaystyle \sum_{n=1}^{\infty}(-1)^na_n$. Now if $~~\displaystyle \lim_{n \to \infty} a_n=0~~$ and $~~(a_n)_{n=1}^{\infty}~~$ is monotone decreasing then by Leibniz test the given series is convergent. But if Leibniz test fails, let $~~\displaystyle \lim_{n \to \infty} a_n=\infty~~$ or $~\neq 0,~~$ what should we conclude about the series $~~\displaystyle \sum_{n=1}^{\infty}(-1)^na_n$. I know, the necessary condition for a positive series to be convergent is limit of the tail tends to zero. But the given series is not positive, and modulus of the a series cannot determine the convergence of the actual series, for this we can take $~~~\displaystyle \sum_{n=1}^{\infty}(-1)^n\frac{1}n.$
So, is there any proof or any discussing paper that, an alternating series will diverge if it fails the Leibniz test? I know the convergence proof (Leibniz test proof), but don't know about the converse.

Please help me to solve this.

  • 4
    $\begingroup$ The convergence of $\sum x_n$ always implies that $x_n \to 0$, not only for series of positive numbers. $\endgroup$
    – Martin R
    Aug 10, 2021 at 19:15
  • $\begingroup$ It can converge if $a_n\to 0$ and $a_n$ is not decreasing, but it doesn't always. For example where it can, if you have the sequence of $a_n$ equal to $\frac{1}{2},1,\frac14,\frac13,\dots.$ Basically, swap the terms $b_n=\frac{1}{n}.$ You can come up with a lot more examples. Given any positive $b_n\to 0$ and any positive $c_n$ with $\sum c_n$ converging, you can define $a_{2n-1}=b_{n},$ $a_{2n}=b_n+c_n,$ then $\sum (-1)^n a_n$ converges but is not decreasing. $\endgroup$ Mar 31 at 6:27
  • $\begingroup$ An example with $a_n\to 0$ not decreasing where the alternating series does not converge is $a_{2n-1}=\frac1n, a_{2n}=\frac{1}{n^2}.$ $\endgroup$ Mar 31 at 6:31
  • $\begingroup$ But if $a_n$ does not converge to zero - whether it converges to a non-zero value, converges to $+\infty,$ or doesn't converge at all - no adding of any signs can make the series converge. $\endgroup$ Mar 31 at 6:42

2 Answers 2


Let's define the partial sum $\displaystyle S_n=\sum\limits_{i=1}^n u_i$.

If the series is convergent then $\ S_n\to \ell\ $, and we get automatically $\ u_n=S_n-S_{n-1}\to \ell-\ell=0$

By contrapositive, if $u_n$ does not converge to $0$, the series diverges.

Of course the conclusion is the same for an alternated series since $|a_n|=|\underbrace{(-1)^na_n}_{u_n}|\to 0$

For the monotonically decreasing part, you can consider this classical counter-example:

$\sum \dfrac{(-1)^n}{\sqrt{n}+(-1)^n}$ is divergent (see a proof here)

while the a priori equivalent series $\sum \dfrac{(-1)^n}{\sqrt{n}}$ is convergent via the Leibniz test.


This is false. First of all:

"I know, the necessary condition for a positive series to be convergent is limit of the tail tends to zero."

For all series, not only for positive.

Your whole idea is false. Nonmonotone decreasing alternating series can be convergent, for example $ \sum_{n=1}^{\infty} (-1)^n\dfrac{\cos^2n}{n^2} $ or $\sum_{n=2}^\infty \dfrac{(-1)^n}{n+(-1)^n}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.