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up vote 10 down vote favorite

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 $0$.

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up vote 28 down vote accepted

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

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1  
So, from what I understand, what matters is that the series is ultimately convergent? – AstlyDichrar 22 hours ago
2  
Exactly, convergence is determined by what ultimately happens. – Oscar Lanzi 22 hours ago
1  
I love it when you can answer questions with tautologies. :) – Mehrdad 15 hours ago
up vote 18 down vote

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

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I have no idea how I didn't think of this, it's the exact same series. – AstlyDichrar 22 hours ago
    
@AstlyDichrar Yes, it is the exact same series ;) – Olivier Oloa 22 hours ago
    
@AstlyDichrar The series is convergent by the alternating test of convergence:en.wikipedia.org/wiki/Alternating_series_test – Olivier Oloa 22 hours ago
    
Indeed, and you could as well just shift indices if you wanted them to start at $1$: $$\sum_{n=1}^\infty (-1)^n \frac{n^2-1}{n^3+1}=\sum_{m=1}^\infty (-1)^{m+1} \frac{(m+1)^2-1}{(m+1)^3+1}.$$ – Ruslan 13 hours ago
up vote 0 down vote

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).

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