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Basically my question is the title:

If have some series $\sum a_n$ which converges and is nonnegative for all its terms, im looking for an example for $a_n$ such that its alternating series $\sum (-1)^n a_n$ is divergent.

I was doing a problem relating to convergence of a power series and looked at the solution and it evoked the alternating series test, which required (1) a decreasing (2) nonnegative sequence (3) whose limit is zero in order for its alternating series of that sequence to converge. The solution did not check if the sequence was decreasing so im a little puzzled and thats why im here.

For anyone interested in the solution, its problem 2(a) at this link:

http://www.northeastern.edu/suciu/MATH3150/MATH3150-fa15-hmw6-solutions.pdf

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    $\begingroup$ There is no such example: the series is absolutely convergent, so you can insert minus signs arbitrarily and still have a convergent series. $\endgroup$ Commented Dec 1, 2016 at 23:53
  • $\begingroup$ I dont know if this is possible: suppose that the series defined in $\Bbb Q$ converges absolutely to some rational number, but the alternating series converges to some irrational number. Then in $\Bbb Q$ the alternating series diverges but the normal series converge. $\endgroup$
    – Masacroso
    Commented Dec 2, 2016 at 0:35
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    $\begingroup$ Your objection to the proof given for 2(a) in the linked pdf is valid. The solution's use of the alternating series test is unjustified. Instead they should simply argue that $\sum |a_n (-R)^n| = \sum a_n R^n$, and since the RHS converges, so does the LHS, which shows that $\sum a_n (-R)^n$ is absolutely convergent, hence convergent. $\endgroup$
    – user169852
    Commented Dec 2, 2016 at 0:38

2 Answers 2

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Notice that

$$|a+b|\le|a|+|b|$$

This is known as triangle rule, and it can be easily seen that it implies

$$|a_1+a_2+a_3+\dots+a_n|\le|a_1|+|a_2|+|a_3|+\dots+|a_n|$$

Putting this into context, it is readily seen that

$$0\le\left|\sum_{n=1}^\infty(-1)^{n+1}a_n\right|\le\sum_{n=1}^\infty\left|(-1)^{n+1}a_n\right|=\sum_{n=1}^\infty a_n=S<\infty$$

So clearly the sum is bounded.

Now, we are left with two cases: the summation bounces or oscillates between at least two values, and thus, it shall never converge but remain bounded as above. Or it will converge. (The third case for any series is to diverge to $\pm\infty$, but we have already shown this not possible)

The term test tells us that

$$\lim_{n\to\infty}a_n=0$$

As should be quite obvious. It then goes to show that our sum cannot oscillate, leaving the only remaining possibility being convergence.

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If a series is absolutely convergent, then it is convergent. The proof can be found here: http://pirate.shu.edu/~wachsmut/ira/numser/proofs/absconv.html

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