Exercise 11 from section 9.3 of Introduction to Real Analysis (Bartle):

Can Dirichlet’s Test be applied to establish the convergence of $$ 1 - \dfrac12 - \dfrac13 + \dfrac14 + \dfrac15 + \dfrac16 - \cdots $$ $\qquad \qquad$ where the number of signs increases by one in each ‘‘block’’? If not, use another method to establish the convergence of this series.

Dirichlet's test cannot be used because the partial sums generated by (1, -1, -1, 1, 1, 1, ...) are not bounded. But we can group the terms of the series in the following way:

$$ 1 - \left(\dfrac12 + \dfrac13\right) + \left(\dfrac14 + \dfrac15 + \dfrac16\right) - \left( \dfrac17 + \dfrac18 + \dfrac19 + \dfrac{1}{10} \right) + \cdots \\ = \sum _{n=1}^{\infty}(-1)^{n+1}a_n $$


$$ (a_n) = \left(1, \left(\dfrac12 + \dfrac13\right), \left(\dfrac14 + \dfrac15 + \dfrac16\right), ... \right) $$

So by Leibniz's test, if the sequence $(a_n)$ is decreasing and $\lim{a_n} = 0$ then the grouped series is convergent. I've shown that since we are grouping terms of the same sign it is sufficient to show the convergence of the grouped series. I've shown that $\lim{a_n} = 0$, but how do I show that $(a_n)$ is decreasing?

  • $\begingroup$ try to generalize: $1/4>1/7+1/30$, $1/5>1/8+1/30$, $1/6>1/9+1/30$ (where $1/30=(1/10)/3$) $\endgroup$ – user8268 Apr 14 '14 at 21:56
  • $\begingroup$ @user8268 I generalized for the case of 1/4 by using a common denominator but it was complicated and I don't know how to extend it to the rest of the terms in each sum as you suggested. $\endgroup$ – Simon Hunt Apr 15 '14 at 3:38

Note that $a_n = \sum_{k=n(n-1)/2+1}^{n(n+1)/2} \frac1k$. In particular, since $\frac1x$ is decreasing, $$ \int_{n(n-1)/2+1}^{n(n+1)/2+1} \frac{dx}x < a_n < \int_{n(n-1)/2}^{n(n+1)/2} \frac{dx}x, $$ or $$ \log\frac{n^2+n+2}{n^2-n+2} < a_n < \log\frac{n+1}{n-1}. $$ In particular, $$ a_n-a_{n+1} > \log\frac{n^2+n+2}{n^2-n+2} - \log\frac{n+2}n = \log\bigg( 1+\frac{2(n-2)}{n^3+n^2+4} \bigg) \ge0 $$ for $n\ge2$.

(In fact, the estimate $|a_{2n-1}-a_{2n}|<\frac1{n^2}$ would suffice to establish convergence, regardless of whether the $a_n$ are decreasing.)

  • $\begingroup$ Regarding your last comment, how do you get the inequality |a_2n-1 - a_2n| < 1/n^2? $\endgroup$ – Simon Hunt Apr 15 '14 at 3:34
  • $\begingroup$ This method plus $\log(1+y)<y$ does it (up to a constant), or something like user8268's comment ... point being, this is an alternate way to approach the problem - concentrating on the size of these terms rather than their sign. $\endgroup$ – Greg Martin Apr 15 '14 at 4:25

Although Dirichlet's test per se does not apply, it seems like a Good Thing to note that the proof of Dirichlet does apply. Sum by parts and you're done.

In more detail: Let $(\epsilon_j)$ be the sequence of plus and minus ones, so we're considering convergence of $$\sum_j\frac{\epsilon_j}{j}.$$Let $\sigma_n=\sum_{j=1}^n\epsilon_j$. Dirchlet does not apply because $\sigma_n$ is not bounded. But it's not hard to see that $$|\sigma_n|\le c\sqrt n.$$(After $N$ "blocks" of ones and minus ones we have $|\sigma_n|\le c N$. But after $N$ blocks we have $n\sim N^2$.)

So sum by parts https://en.wikipedia.org/wiki/Summation_by_parts :

$$\sum_{j=1}^n\frac{\epsilon_j}{j}=\frac{\sigma_n}{n}-\sum_{j=1}^{n-1}\sigma_j \left(\frac1{j+1}-\frac1j\right).$$Since $\sqrt n/n\to0$ and $\sum\sqrt j/j^2<\infty$ the sum converges.

Moral Proofs of theorems are even better than theorems.

  • $\begingroup$ I am trying to understand your argument. What is '$c$' in $|\sigma_n|\le c\sqrt n.$? $\endgroup$ – Error 404 Mar 2 '17 at 16:36
  • 1
    $\begingroup$ @VikrantDesai $c$ is some constant, the value of which doesn't matter. Say $T_k=1+2+\dots+k$. Then $T_k\sim k^2/2$. It's clear that $|\sigma_{T_k}|\le ck$. Given $n$, choose $k$ with $T_k\le n<T_{k+1}$. Then $|\sigma_n-\sigma_{T_k}|\le k$, hence $|\sigma_n|\le k+ck\le c\sqrt n$. (Here we used the standard convention that the letter "$c$" refers to different constants in each occurrence.) $\endgroup$ – David C. Ullrich Mar 4 '17 at 15:20
  • $\begingroup$ (+1)Thanks for your reply. If we take $c=2$ or greater than $2$ then won't it be safe? or $c$ depends on $N$, the number of blocks of ones and minus ones? Sorry for my late response. $\endgroup$ – Error 404 Mar 6 '17 at 8:44

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.