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I'm doing a question for homework, and I am required to use the Comparison Test to test for convergence.

The series in question is:

$$ \sum_{n=0}^\infty \frac {n-1}{{(n+2)}^3} $$

The series I would like to compare it to is:

$$ \sum_{n=1}^\infty \frac 1{{n}^2} $$

My reasoning for wanting to use this series is because it is a convergent p-series that is larger than the series I am trying to test, but I am unsure if I can use it as comparison due to its divergence to $ \infty $ at n=0. I recall that with limits what is important is its behaviour as $ n\to\infty$, so does that also apply for comparing series?

Edit: Thank you to all who answered! Each one contributed to my understanding. :)

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You can even compare a series that begins with $\,n=1\,$ with another one that begins with $\,n=10^{10^{10}}\,$ (positive series, of course). – DonAntonio Sep 23 '12 at 16:16
up vote 8 down vote accepted

We have:

$$ \sum_{n=0}^N \frac{n-1}{{(n+2)}^3} = \frac{0-1}{{(0+2)}^3} + \sum_{n=1}^N \frac {n-1}{{(n+2)}^3} = -\frac{1}{8}+ \sum_{n=1}^N \frac {n-1}{{(n+2)}^3} $$

Hence, the series starting at $n=0$ and that starting at $n=1$ converge together or diverge together. This is true for any starting index, since the difference is always finite.

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Thank you for the specific answer! Really helped illustrate the point. – quelledanielle Sep 23 '12 at 16:21
@quelledanielle Glad it helped! – Ayman Hourieh Sep 23 '12 at 16:22

Yep, you can compare them. If you want to prove this rigorously, plug in n+1 for n in the second sigma function, then apply the comparison test.

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Yes you can, because for comparison the first $N$ terms don't matter, for any fixed $N$.

Alternately, in the first expression, you could let $m=n+1$, and then your first series is $\displaystyle\sum_{m=1}^\infty \frac{m-2}{(m+1)^2}$.

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$$ \sum_{n=0}^\infty \frac {n-1}{(n+2)^3} = \underbrace{\frac{0-1}{(0+2)^3}}_{\text{The }n=0\text{ term}} + \underbrace{\sum_{n=1}^\infty \frac {n-1}{(n+2)^3}}_{\text{The other terms}} $$

The question of whether the series on the left converges is the same as the question of whether the series on the right, with only the terms other than that in which $n=0$, converges. You can compare that with the other series.

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Because adding or subtracting finite quantity of summands preserve series convergence, you can separate term with $n=0$ and then apply comparison test

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First, note that $\sum\frac{1}{n^2}$ converges. It is a $p$-series with $p = 2 > 1$.

Second, yes: the asymptotic behavior of the series is what matters. If there is an index $N$ beyond which $a_k < b_k$, and $\sum b_k$ converges, then you are guaranteed that $\sum a_k$ converges. Conversely, if $\sum a_k$ diverges, then you are guaranteed that $\sum b_k$ diverges. This is because below any finite index, the sum is finite and therefore does not impact the convergence of the series.

Third: Here's a hint for solving your problem. You can't easily compare these series since the denominators are different. Try reindexing the series so that the denominators are the same. For example, put $k = n + 2$ in the first series, so that it becomes $$\sum_{k = 2}^\infty \frac{k - 3}{k^3}.$$ Can you manipulate this to profitably compare it to $\sum\frac{1}{k^p}$ for some $p$?

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The third part seems very helpful to improve the clarity of my comparison, thank you! – quelledanielle Sep 23 '12 at 16:32
You're quite welcome, @quelledanielle. – Neal Sep 23 '12 at 17:56

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