# If I can't use Alt. Series Conv. test, how do I determine if it converges or diverges?

Stumped on this absolute convergence problem!
(Converge conditionally, absolutely, or diverges)

$$\sum_{n=1}^\infty \frac{(-1)^n n^2}{(n+3)^2}$$

First, I tested for absolute convergence, and the fraction reduced to $\frac{n^2}{n^2}=1$ .....So, it does not converge absolutely.

Next, to use the alternating series test on the original series, you need to make sure the magnitude of the terms are decreasing. Well, they are not:

$$\sum_{n=1}^\infty \frac{(-1)^n n^2}{(n+3)^2} = -\frac{1}{16} + \frac{4}{25} - \frac{9}{36} + .....$$

So, without using the alternating series test, how do I determine if the orig. series converges or diverges?

• For which values of $n$ is $\frac{n^2}{(n+3)^2}$ smaller than $\frac15$? – user228113 Mar 22 '16 at 14:02
• Do the terms tend to $0$? – David Mitra Mar 22 '16 at 14:04
• Use the divergence test. – Brian M. Scott Mar 22 '16 at 14:05
• Why are you not allowed to use Alt Series test if this is an alt series? Just curious – imranfat Mar 22 '16 at 14:07
• @imranfat Because it doesn't apply. The alternating series test requires successive terms to tend to $0$. (That is, it requires the series to be "possibly convergent".) – Patrick Stevens Mar 22 '16 at 14:08

$$\frac{n^2}{(n+3)^2}\xrightarrow[n\to\infty]{}1\implies (-1)^n\frac{n^2}{(n+3)^2}\rlap{\;\;\;\;/}\xrightarrow[n\to\infty]{}0$$

• This answer is one part of the Alt Series Test which the OP cannot use. – imranfat Mar 22 '16 at 14:07
• @imranfat Thank you. I beg to difer, though: this is the basic necessary condition for any convergent series, and thus any sequence not converging to zero its series cannot converge. The Alt. Series test requires the convergence to zero to be monotonic. – DonAntonio Mar 22 '16 at 14:09
• Yes, that is true and I would use the same argument to show that this series does not converge. But it is also one of the 2 conditions for the alt series test to converge (The other one being that it has to be monotonic decreasing) – imranfat Mar 22 '16 at 14:12
• @imranfat Again, I don't think convergence to zero can be a special condition for the A.S. test: it applies to all series! Of course, any convergence test will imply first that the series' sequence must converge to zero, otherwise there's no point in applying the test. – DonAntonio Mar 22 '16 at 14:15
• This condition shows up in two tests, the alternating series test and the test for divergence. Even though the alternating series test can't be applied, the divergence test can be applied to show divergence. – Michael Burr Mar 22 '16 at 14:19

Indeed, Clement's comment below is correct. This sum is not convergent and has no well-defined value!

The method below is akin to resummation but still very problematic. For this sum, the Cesaro sum converges to 0 and Abel's resummation method gives $\frac{1}{8} \left(-26+6 \pi ^2-48 \log (2)\right)\approx -0.00667$. Both differ from the value computed below.

-- I'm leaving the equation below as a reference for Clement's comment, Beware --

One approach is to combine odd and even terms. This gives a sequence with positive terms, for which convergence is easy to determine. Thus, we have $$\sum_{n=1}^\infty \frac{(-1)^n n^2}{(n+3)^2} = \sum_{k=0}^\infty \left( -\frac{(2k+1)^2}{(2k+4)^2} + \frac{(2k+2)^2}{(2k+5)^2} \right) = \sum_{k=0}^\infty \frac{39+72k+24 k^2}{4(k+2)^2(2k+5)^2} = \frac{3 \pi^2 - 12 \ln (4)-11}{4} \approx 0.493$$

• But the series does not converge. Its general term does not even go to zero, which is the most basic necessary condition... You cannot combine terms as you want, otherwise the series $\sum_n (-1)^n$ would converge, for instance. – Clement C. Mar 22 '16 at 14:20