# Leibniz's test - converting to a positive number because first element is positive?

At the last lecture I had the teacher stated that if a given series is a series of the form: $$\sum_{n=0}^\infty (-1)^na_n$$ that satisfies Leibniz's test, and the first element of the series (with $n=0$) is positive, then the series converge to a positive number.
I really didn't understand where this comes from. Can someone please explain?

Would the series converge to a negative number if the first one was negative?

• What's the definition of a Leibniz series in this context?
– dxiv
Jan 2, 2017 at 23:17
• @dxiv I'm sorry, I thought it's a common thing people say and not just among us. Edited the question :) Jan 2, 2017 at 23:31
• Hint: the limit lies between any two consecutive partial sums. In particular, between $a_0$ and $a_0-a_1$.
– dxiv
Jan 2, 2017 at 23:34
• @dxiv Why is this true? Jan 2, 2017 at 23:37
• It follows from the usual proof of the alternating series (Leibniz) test.
– dxiv
Jan 2, 2017 at 23:39

Would the series converge to a negative number if the first one was negative?

Possible, assuming you have $(a_n)$ satisfying Leibniz' test (i.e. $(|a_n|)$ monotonically decreasing and converging to $0$ and all $a_n$ of the same signum) with $a_0>0$, then the sequence $(-a_n)$ would also satisfy Leibniz' test and you have

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

On the other hand, with $s := \sum_{n=0}^\infty (-1)^{n}a_n$ you could set $b_0 = -\frac{s}{2}$ and for $n>0$ set $b_n = a_{n-1}$, then

$\sum_{n=0}^\infty (-1)^{n} b_n = \frac{s}{2} > 0$ but $b_0 < 0$.

Hint: the limit lies between any two consecutive partial sums. In particular, between $a_0$ and $a_0−a_1$. It follows from the usual proof of the alternating series (Leibniz) test.