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Maybe I am reading the mathese wrong but according to my book: If $a_n$ is positive and decreasing and $\,\displaystyle\lim_{n\to\infty} a_n = 0,\,$ then the alternating series converges.

So for example if I have $\,\displaystyle\sum_1^\infty \frac{1}{n},\,$ I know that diverges so $\,\displaystyle\sum_1^\infty \frac{(-1)^n}{n}\,$ should also diverge but according to the Leibniz test it is decreasing and it's limit is zero so it should converge. What does this mean?

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    $\begingroup$ Why do you think the alternating series "should also diverge"? The answers in your earlier question are suggesting that since the harmonic series diverges, the alternating harmonic series cannot converge absolutely (not that it can't converge). $\endgroup$ – anon Jul 22 '13 at 2:43
  • $\begingroup$ The series converges to $-\log 2$. However, the series would be a terrible way to try to approximate $\log 2$ because convergence is quite slow. We need to add up $2000$ terms to get $3$ decimal place accuracy! $\endgroup$ – André Nicolas Jul 22 '13 at 2:51
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The alternating series $$\sum_{n = 1}^\infty \frac{(-1)^n}{n}$$ converges, as you note, conditionally, and by the Leiniz test. It does not converge absolutely. Please note that

$$\sum_{n = 1}^\infty \frac{(-1)^n}{n} \neq \sum_{n = 1}^\infty \frac 1n$$

Alternating series are not equivalent to their non-alternating counterparts. The alternating series test, aka the Leibniz Test, is precisely for series like this: series that do not converge absolutely, but converge nonetheless. The fact that alternating terms are negative negates the divergence of the sum in the case of the absolute term.

Absolute convergence simply means the series consisting of the absolute values of the terms converges. It does not mean "definitely" converges. A conditionally convergent series still very much converges, for a different reason, but converges just the same.

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    $\begingroup$ @John The test says the series with with the $(-1)^n$ converges, not the series without the $(-1)^n$. $\endgroup$ – Pedro Tamaroff Jul 22 '13 at 2:44
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    $\begingroup$ @John A series $\sum a_k$ is absolutely convergent if $\sum |a_k|$ converges. It is conditionally convergent if $\sum a_k$ converges but $\sum |a_k|$ diverges. $\endgroup$ – Pedro Tamaroff Jul 22 '13 at 2:50
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    $\begingroup$ It's the definition: a series $\sum a_n$ is said to converge if $\sum_{n=0}^N a_n$ has a limit when $N$ goes to $\infty$. A series $\sum a_n$ is said to converge absolutely if the series of the absolute values, $\sum |a_n|$, converges (note that this is stronger: "absolutely convergent" implies "convergent"). If a series converges, but does not converge absolutely, it is said to converge conditionally. $\endgroup$ – Clement C. Jul 22 '13 at 2:58
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    $\begingroup$ It will begin to sink in. It does get very confusing: so many tests, conditions, etc. can be overwhelming when first learning these. I just usually check first for absolute convergence. If the series with the absolute values of terms converges, you are done. If not, we don't know yet. If it passes the alternating test, then you can say, yes, converges. $\endgroup$ – Namaste Jul 22 '13 at 2:59
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    $\begingroup$ Not quite John: if a series converges with absolute values, it converges absolutely. If it does not converge with absolute values, it might still converge, but conditionally, as with this series. $\endgroup$ – Namaste Jul 22 '13 at 3:04
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Everything in the question makes sense EXCEPT "so $\sum_1^\infty \frac{(-1)^n}{n}$ should also diverge". That's wrong. It is NOT TRUE that if $\sum_{n=1}^\infty a_n$ converges, then so does $\sum_{n=1}^\infty |a_n|$.

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  • $\begingroup$ Regardless the limit is the same in both cases. $\endgroup$ – John Jul 22 '13 at 2:46
  • $\begingroup$ If you mean $\lim_{n\to\infty} a_n$, it is true that that is the same as $\lim_{n\to\infty} |a_n|$ in cases in which $\sum_{n=1}^\infty a_n$ converges. $\endgroup$ – Michael Hardy Jul 22 '13 at 2:51
  • $\begingroup$ @John: what do you mean? $\sum \frac{1}{2^n}$ and $\sum \frac{(-1)^n}{2^n}$ both converge, but $\sum_{n=0}^\infty \frac{1}{2^n}=2$, while $\sum_{n=0}^\infty \frac{(-1)^n}{2^n}=\frac{2}{3}$. $\endgroup$ – Clement C. Jul 22 '13 at 2:55
  • $\begingroup$ @MichaelHardy: +1 thanks for the edit you did for my answer. Thanks for the time. :) $\endgroup$ – mrs Jul 22 '13 at 18:33

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