Tuesday, 2 July 2013

Offset Alternating Series



I have the following alternating series that I would like to determine whether it is absolutely convergent, conditionally convergent, or divergent:



n=11+2(1)nn




I have applied some tests and I find it reasonable to conclude that it is divergent.



As a sum of two series:
n=11n+n=12(1)nn



I believe a convergent series when added to a divergent series, results in a divergent series. If this isn't a fact then I would still be left to say that it is inconclusive.



Using the Alternating Series Test, with:
an=1+2(1)nn

although this isn't of 'proper form' n=1(1)nan the limit of an does approach zero as n. As for monotonically decreasing, the limit of the ratio of absolute terms is divergent for n even and inconclusive for n odd, which has me concluding divergent by The Ratio Test as well as not monotonically decreasing, where:



lim



Am I on the right track here? Am I making any really improper assumptions? Was there a better way to go about with the proof?



Thanks!


Answer



To use the alternating series test, you need to rewrite the series first; you cannot use it directly with the a_n you are given.




Define b_n as follows: when n is even, b_n = \frac{3}{n}; when n is odd, b_n = \frac{1}{n}. Then (-1)^nb_n = a_n, so you can apply the alternating test to your original series by writing the series as \sum (-1)^n b_n.



Unfortunately, the alternating series test does not apply: remember that you need not only that \lim\limits_{n\to\infty} b_n=0, but also that b_{n+1}\lt b_n for all sufficiently large n. However, for n odd, we have b_n\lt b_{n+1}, since \frac{1}{n}\lt \frac{3}{n+1} if and only if n+1\lt 3n, which holds for all n. So the alternating series test cannot be used.



The Ratio Test is likewise inconclusive: the limit does not exist. To see this, note that for n odd, we have
\left|\frac{a_{n+1}}{a_n}\right| = \frac{\quad\frac{3}{n+1}\quad}{\frac{1}{n}} = \frac{3n}{n+1}
which converges to 3 as n\to\infty; whereas for n even, we have
\left|\frac{a_{n+1}}{a_n}\right| = \frac{\quad\frac{1}{n+1}\quad}{\frac{3}{n}} = \frac{n}{3(n+1)}
which converges to \frac{1}{3}. Since the sequence of odd terms and the sequence of even terms of |\frac{a_{n+1}}{a_n}| converge to different things,
\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|

does not exist. So the Ratio Test is inconclusive.



As to your first method, it's fine: indeed, the sum of a divergent series and a convergent series must diverge. To see this, note that if \sum r_n and \sum(r_n+s_n) both converge, then so does \sum ((r_n+s_n)-r_n) = \sum s_n, since the difference of convergent series converges. Thus, if any two of \sum r_n, \sum s_n, and \sum (r_n+s_n) converge, then so does the third, and in particular if any one of the series diverges, then at least two must diverge.


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