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I'm reading the following set of notes on Taylor series and big O-notation, written by a professor at Columbia: http://www.math.columbia.edu/~nironi/taylor2.pdf. He repeatedly refers to what he calls "limit comparison", by which he means the theorem that for $a_n, b_n$ sequences of positive real numbers such that $b_n \to C>0$, we have that $\sum a_n <\infty$ iff $\sum a_n b_n < \infty$. On page 13, he is manipulating a series using O notation, and he ends up with $$\sum\limits_{n=1}^{\infty}(-1)^n\left(\frac{1}{n}\right)\frac{\frac{7}{12}+O(1/n^2)}{-\frac{1}{6}+O(1/n^2)},$$at which point he says "At this point we might be tempted to use limit comparison and conclude that the series is convergent; but limit comparison cannot be applied to an alternating series. Instead of using limit comparison we try to separate the part of the series that converges but not absolutely from the part that converges absolutely."

Now, in reference to the theorem he calls "limit comparison", can't we strengthen this theorem to say that if $a_n$ and $b_n$ are sequences of reals, and $b_n \to C \neq 0$, then $\sum a_n <\infty \iff \sum a_n b_n < \infty$? If not, what is a relevant counter-example? And if so, can't we use this to conclude that the above series converges?

My second question is that I do not understand the algebraic manipulation he does directly after the sentence quoted above. I suppose he is trying to "separate the part of the series that converges but not absolutely from the part that converges absolutely", but I don't know what this means, or what he is doing. This is all on page 13.

Thanks for your help.

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  • $\begingroup$ Thank you for the edit, @TedShifrin, I am sorry for the typo. $\endgroup$ – Eric Auld Jun 30 '13 at 1:58
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So here's all that's going on with that series. Remember that $O(1/n^2)$ refers to a term that is bounded by a multiple of $1/n^2$.

Recall that $\dfrac1{1+x} = 1-x+x^2 + \dots$ for $|x|<1$. In particular, $\dfrac1{1+O(1/n^2)} = 1+ O(1/n^2)$. So \begin{align*} \frac{\frac7{12} + O(1/n^2)}{-\frac16+O(1/n^2)} &= \frac{\frac7{12} + O(1/n^2)}{-\frac16\big(1+O(1/n^2)\big)} = \big(-\tfrac72 + O(1/n^2)\big)\big(1+O(1/n^2)\big) \\ &= -\tfrac72 + O(1/n^2)\,, \end{align*} the last step following by distributing.

This means that his original series is of the form $$\sum \left((-1)^{n+1}\frac72\cdot\frac 1n + O(1/n^3)\right)\,.$$ This is a sum of two convergent series, hence convergent. The first, however, is conditionally convergent, whereas the second is absolutely convergent. This means that the entire series is conditionally convergent. (He comments that you could see this directly by removing the $(-1)^n$ and then using limit comparison with the harmonic series.)

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  • $\begingroup$ Thank you so much for parsing this for me! I understand now. $\endgroup$ – Eric Auld Jun 30 '13 at 4:03
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Let $a_n=\frac{(-1)^n}{n^{1/4}}$, and let $b_n=1+\frac{(-1)^n}{n^{1/4}}$. Then $\sum a_n$ converges, and $(b_n)$ has limit $1$.

We have $a_nb_n=a_n +\frac{1}{n^{1/2}}$. Convergence fails for $\sum a_nb_n$.

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  • $\begingroup$ Thank you! This totally answers my first question. $\endgroup$ – Eric Auld Jun 30 '13 at 1:49
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    $\begingroup$ For the other part, I do not know the context. But I think there may be a typo, want $O(1/n^2)$. $\endgroup$ – André Nicolas Jun 30 '13 at 2:00
  • $\begingroup$ Yes, the typo was mine, regrettably. Ted corrected it. $\endgroup$ – Eric Auld Jun 30 '13 at 2:00
  • $\begingroup$ I guessed right! Then the expression can be replaced by $-\frac{7}{2}+O(1/n^2)$, giving convergence. In a sense it is the opposite of what I did in my answer, I made the errors "big." $\endgroup$ – André Nicolas Jun 30 '13 at 2:02
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    $\begingroup$ I just addressed that. See if you're content :) $\endgroup$ – Ted Shifrin Jun 30 '13 at 2:06

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