0

Is there a way to bound the sum $$\sum_{k=1}^\infty \frac{a_k}{k}$$ given that you already know the value of $$\sum_{k=1}^\infty a_k$$

jimjim
  • 9,675
CogitoErgoCogitoSum
  • 3,521
  • 8
    $$\sum_{k=1}^\infty \frac{a_k}{k} \leq \sum_{k=1}^\infty a_k$$ – Git Gud Mar 02 '13 at 21:48
  • 9
    Are the $a_k$ nonnegative? – Julien Mar 02 '13 at 21:55
  • 4
    I think there is no way to determine bound first sum using last sum. (I think $a_k=(-1)^{b_k}/\ln(1+k)$ is appropriate example.) – Hanul Jeon Mar 02 '13 at 23:01
  • 2
    Or consider $a_k = \begin{cases} M,&\text{when $k=n$} \0&\text{otherwise}\end{cases}$. – MJD Mar 02 '13 at 23:40
  • GitGud, sorry, I guess I should have specified that the most trivial case is not what Im asking. I was hoping to get a little thought put into this. My bad, once again, for choosing this web site. A tighter bound than that would be excellent.

    julien, the reason I used $a_k$ was to denote an arbitrary sequence of values. Im sorry I need to specify this, too. I thought it was implied by the lack of further information.

    MJD Im not entirely sure what youre getting at. You are creating a sequence of one term?

     – CogitoErgoCogitoSum Mar 03 '13 at 01:46
  • If we knew the given sum was absolutely convergent then I believe you could apply Dirichlet's test, http://en.wikipedia.org/wiki/Dirichlet%27s_test – James S. Cook Mar 03 '13 at 05:21
  • $\sum_{k=1}^{n} \frac{a_k}{k} = \dfrac{\sum_{k=1}^{n}a_k}{n!} \le \dfrac{\text{any upper bound on $\sum_{k=1}^{n}a_k$ }}{n!} $ but this bounds the partial sum. – Rustyn Mar 04 '13 at 21:55
  • 14
    Cogito, if you so dislike this website, you know no one is forcing you to come here. Also, if you want others to put some thought into your question, you might try putting some thought into its presentation yourself. And try putting some thought into the answers of tetori and MJD. – Gerry Myerson Mar 04 '13 at 22:28
  • 2
    @CogitoErgoCogitoSum: it is not necessarily clear what is meant by $a_k$. Often people forget to mention conditions such as $a_k\ge0$, or people inadvertently assume such conditions. Although the question is trivial if $a_k\ge0$, it would avoid confusion to mention that this is not the case. Certainly, I see no reason to be upset by such inadvertent assumptions, and definitely not by such questions for clarification. – robjohn Mar 05 '13 at 10:48
  • 1
    @JamesS.Cook: Dirichlet's test applies even if the series is not absolutely convergent. In fact, I have noticed it being most useful when the series is not absolutely convergent. – robjohn Mar 05 '13 at 10:53
  • similar, sort of related : http://math.stackexchange.com/questions/1766/if-sum-n-1-infty-a-n3-converges-does-sum-n-1-infty-fraca?rq=1 – jimjim Mar 05 '13 at 20:18
  • I choose math.se because you guys are more competent with deeper knowledge of math. I end up going to yahoo answers because they end up being more helpful and cooperative, and dont need every little statement defined explained for them. – CogitoErgoCogitoSum Mar 05 '13 at 22:50
  • 1
    @robjohn thanks for the note, I'll try to be less greedy next time. – James S. Cook Mar 06 '13 at 03:51
  • @JamesS.Cook: greedy? I must be missing something. – robjohn Mar 06 '13 at 04:21
  • @robjohn sorry, I mean to say I have assumed more than is required of the Dirichlet Test. – James S. Cook Mar 07 '13 at 23:13

1 Answers1

7

Using the Dirichlet Convergence Test, if $$ \left|\,\sum_{k=1}^na_k\,\right|\le A $$ for all $n$, and if $\displaystyle\lim_{n\to\infty}b_n=0$ and $$ \sum_{k=1}^\infty\left|\,b_{k+1}-b_k\,\right|=B $$ Then $$ \left|\,\sum_{k=1}^na_kb_k\,\right|\le AB $$ Thus, $$ \left|\,\sum_{k=1}^\infty\frac{a_k}{k}\,\right|\le\sup_{n\ge1}\,\left|\,\sum_{k=1}^na_k\,\right| $$ but I don't see an obvious way to get a better bound.

Consider the sequence $$ a_k=\frac{(-1)^{k+1}}{\left\lfloor{\Large\frac{k+1}2}\right\rfloor} $$ Then the series $$ \begin{align} \sum_{k=1}^\infty a_k &=\sum_{k=1}^\infty\frac1k(1-1)\\[6pt] &=0 \end{align} $$ yet $$ \begin{align} \sum_{k=1}^\infty\frac{a_k}{k} &=\sum_{k=1}^\infty\frac1k\left(\frac1{2k-1}-\frac1{2k}\right)\\ &=\sum_{k=1}^\infty\left(\frac2{2k-1}-\frac2{2k}-\frac1{2k^2}\right)\\ &=2\log(2)-\frac{\pi^2}{12}\\[10pt] &\stackrel.=0.563827327695778 \end{align} $$ Thus, there can be no constant, $c$, so that $$ \left|\,\sum_{k=1}^\infty\frac{a_k}{k}\,\right|\le c\,\left|\,\sum_{k=1}^\infty a_k\,\right| $$

robjohn
  • 345,667
  • How does that prove that there can be no constant $c$ whatsoever? Its fine if $\sum \frac{a_k}{k} \ge \sum a_k$. Im just looking for a way to bound the relationship. And youre assuming that the relationship HAS to be proportional involving a scalar factor... rather than, say, a scalar addition. Furthermore, dont you contradict yourself? The first half of your argument is proving $\sum \frac{a_k}{k} \le \sum a_k$, but then you go on to give an example of the reverse inequality, thus disproving the first half of your argument. – CogitoErgoCogitoSum Nov 27 '14 at 22:02
  • @CogitoErgoCogitoSum: the sum on the right side is $0$; the sum on the left side is positive. There is no constant, $c$, so that $2\log(2)-\frac{\pi^2}{12}\le c\cdot0$. – robjohn Nov 27 '14 at 22:48
  • I made about three or four points. Youre going to focus on one? Youre barely touched on what I said. For example, why must the bound be a scalar constant proportionality? Why cant we add or subtract a scalar value? Is there NO way to bound? Or simply no way to bound with a scalar factor? Youve only proven that it can be done with one assumption of the form it ought to take. – CogitoErgoCogitoSum Nov 28 '14 at 01:53
  • 1
    @CogitoErgoCogitoSum: Using the sequence I give above, the sequence $b_1=d+ca_1$ and $b_k=ca_k$ for $k\gt1$ is a sequence so that $\sum\limits_{k=1}^\infty b_k=d$ and yet $\sum\limits_{k=1}^\infty\frac{b_k}{k} =d+c\left(2\log(2)-\frac{\pi^2}{12}\right)$. Adjusting $d$ and $c$, we can get any two values for $\sum\limits_{k=1}^\infty b_k$ and $\sum\limits_{k=1}^\infty\frac{b_k}{k}$. – robjohn Nov 28 '14 at 07:45