I have an alternate series which I want to test for convergence or divergence. The series is as follows:

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

I know how to test this for convergence, but the first term is $0$ and so "$n+1$" terms are not allways smaller than $n$ terms. I have seen the answer and the series is convergent (although not absolutely, but I knew that from testing $\sum_{n=1}^\infty \frac{n^2-1}{n^3+1}$ in a previous exercise), can I just "throw out" the $0$ and say it doesn't matter in the grand scheme of things? The terms of the series tend to $0$, so the conditions for convergence in alternate series are satisfied except for that nasty $0$.

  • $\begingroup$ "The terms of the series tend to 0, so the conditions for convergence in alternate series are satisfied except for that nasty 0" And except for the crucial condition that the terms are decreasing in absolute value. Did you check they are? $\endgroup$ – Did Dec 31 '16 at 23:14
  • 1
    $\begingroup$ ((FWIW, the votes on this page seem rather irrational.)) $\endgroup$ – Did Dec 31 '16 at 23:18
  • $\begingroup$ let us call this a multivariate (two-input function where x is the end of the sum and n is the start) function f(n,x). Why wouldn't f(1, infinity) = f(1,1) + f(2,infinity) = f(2,infinity) $\endgroup$ – The Great Duck Jan 1 '17 at 4:05

You can remove a finite number of terms and not affect convergence.

  • 2
    $\begingroup$ So, from what I understand, what matters is that the series is ultimately convergent? $\endgroup$ – AstlyDichrar Dec 31 '16 at 0:50
  • 2
    $\begingroup$ Exactly, convergence is determined by what ultimately happens. $\endgroup$ – Oscar Lanzi Dec 31 '16 at 0:54
  • 3
    $\begingroup$ I love it when you can answer questions with tautologies. :) $\endgroup$ – Mehrdad Dec 31 '16 at 7:55
  • 1
    $\begingroup$ Indeed, questions with answers which are the answers to the questions themselves are best. $\endgroup$ – Mateen Ulhaq Jan 1 '17 at 1:53
  • $\begingroup$ google.com/url?sa=t&source=web&rct=j&url=https://… $\endgroup$ – Oscar Lanzi Jul 17 '17 at 19:30

Observe that your series just rewrites $$ \sum_{n=1}^\infty (-1)^n \frac{n^2-1}{n^3+1}=\sum_{n=\color{red}{2}}^\infty (-1)^n \frac{n^2-1}{n^3+1}. $$

  • $\begingroup$ I have no idea how I didn't think of this, it's the exact same series. $\endgroup$ – AstlyDichrar Dec 31 '16 at 0:49
  • $\begingroup$ @AstlyDichrar Yes, it is the exact same series ;) $\endgroup$ – Olivier Oloa Dec 31 '16 at 0:51
  • $\begingroup$ @AstlyDichrar The series is convergent by the alternating test of convergence:en.wikipedia.org/wiki/Alternating_series_test $\endgroup$ – Olivier Oloa Dec 31 '16 at 0:52
  • $\begingroup$ Indeed, and you could as well just shift indices if you wanted them to start at $1$: $$\sum_{n=1}^\infty (-1)^n \frac{n^2-1}{n^3+1}=\sum_{m=1}^\infty (-1)^{m+1} \frac{(m+1)^2-1}{(m+1)^3+1}.$$ $\endgroup$ – Ruslan Dec 31 '16 at 9:40

If your sequence is $a_n$, you could test the series for the sequence for $b_0 = 1$ (or $-1$) and $b_n = a_n$ for $n > 1$ for convergence.

You can "cleanly" apply the convergence test to $b_n$, and I will leave as an exercise relating that to the series for $a_n$ (it is not hard). Because this hack is so trivial we usually just apply it "sloppily," but you are definitely doing the right thing by asking how to do it properly.

In general when testing for convergence for any series you can do any arbitrary manipulation to the first $N$ terms you want (i.e. you "ignore" them, whatever that needs to mean).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.