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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?

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

1 Answer 1

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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 $b_0 < 0$.

The other part of your question was already answered by dxiv in the comments:

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.

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