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This sequence is a rearrangement of the series $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}...$. Note that at this point in the text we do not have any theorem about the convergence of rearrangements.

Let $\{s_n\}$ be the sequence of partials sums of the series then for $n \ge 0$

$s_{3(n+1)} = \sum ^n _ {k=0} \frac{1}{4k+1} + \frac{1}{4k+3} - \frac{2}{4k+4}$

We can view it as the sequence(on $n$) of partials sums of

$\sum_0 a_n = \sum_0 \frac{1}{4n+1} + \frac{1}{4n+3} - \frac{2}{4n+4}$

Where $|a_n| = a_n = \frac{1}{4n+4}\{\frac{3}{4n+1}+\frac{1}{4n+3}\} \le \frac{1}{4n^2}$.

By the comparison test $s_{3(n+1)}$ converges to some real $\alpha$.

But $s_{3(n+1)+1} = s_{3(n+1)}+ \frac{1}{4n+5}$ and $s_{3(n+1)+2} = s_{3(n+1)}+ \frac{1}{4n+5}+\frac{1}{4n+7} $hence we have a partition of $\{s_n\}$ into subsequences which tend to $\alpha$ and this implies $s_n \rightarrow \alpha$.

Is my proof correct? Any alternative solutions are appreciated.

B. Freitas
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4 Answers4

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By the Riemann-Dini theorem, we may take any series that is conditionally convergent but not absolutely convergent and rearrange it in order to get a series that converges to $\alpha$, for any $\alpha\in\mathbb{R}$.

In our case: $$\begin{eqnarray*} \sum_{k\geq 0}\left(\frac{1}{4k+1}+\frac{1}{4k+3}-\frac{1}{2k+2}\right)&=&\sum_{k\geq 0}\int_{0}^{1}\left(x^{4k}+x^{4k+2}-2 x^{4k+3}\right)\,dx\\&=&\int_{0}^{1}\frac{1+x^2-2x^3}{1-x^4}\,dx\\&=&\frac{3}{2}\log 2.\end{eqnarray*} $$

We may notice that we know in advance that the LHS is converging, since: $$ \frac{1}{4k+1}+\frac{1}{4k+3}-\frac{1}{2k+2} = \frac{8k+5}{(4k+1)(4k+3)(2k+2)}=O\left(\frac{1}{k^2}\right).$$

Convergence also follows from Dirichlet's test, since the sequence $1,1,-2,1,1,-2,\ldots$ has bounded partial sums while the sequence $\frac{1}{1},\frac{1}{3},\frac{1}{4},\frac{1}{5},\frac{1}{7},\frac{1}{8},\ldots$ decreases to zero.

Jack D'Aurizio
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  • Does this deal with the convergence issue (which is the OP's problem)? – Did Jul 21 '15 at 21:30
  • Yeah, this is not what I'm looking for. I was hoping for a smaller argument without theorems specific to rearrangements(since Rudin has not introduced them yet). – B. Freitas Jul 21 '15 at 21:32
  • Relax, people. A single line explaining convergence has yet been added. – Jack D'Aurizio Jul 21 '15 at 21:33
  • The convergence of $s_{3n}$ is easier with comparison imo. Do you know a way for proving that the initial series is convergent directly? – B. Freitas Jul 21 '15 at 21:37
  • @B.Freitas: added that, too. Dirichlet's test is the key. – Jack D'Aurizio Jul 21 '15 at 21:38
  • Actually, "we know in advance that the LHS is converging, since" the estimate you wrote holds and because the supplementary step in the OP's question holds. – Did Jul 21 '15 at 21:42
  • @Did: I do not understand what you are criticizing. What is wrong with the above lines? – Jack D'Aurizio Jul 21 '15 at 21:44
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    I think he means that the first part of your answer is enough only if the less obvious part of my argument holds. Dirichlet's test works perfectly though. – B. Freitas Jul 21 '15 at 21:47
  • Just what I already wrote: in the OP's notations, $(s_{3n})$ converges, hence a supplementary argument is necessary to deduce that $(s_n)$ converges. Said argument is not in your answer but it is in the question so everything is cool, I guess. (Re Dirichlet's test: yes it is pretty, but calling it "the" argument seems misleading.) – Did Jul 21 '15 at 21:47
  • Oh, ok, I did not mention that the convergence of $s_{3n}$ implies the convergence of $s_n$ since the general term is infinitesimal, but that is pretty obvious and yet mentioned by the OP. – Jack D'Aurizio Jul 21 '15 at 21:50
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Showing Convergence

Breaking the series into chunks of $3$ terms, which is okay since the terms tend to $0$, we get $$ \begin{align} \sum_{k=0}^\infty\left[\frac1{4k+1}+\frac1{4k+3}-\frac2{4k+4}\right] &=\sum_{k=0}^\infty\left[\left(\frac1{4k+1}-\frac1{4k+4}\right)+\left(\frac1{4k+3}-\frac1{4k+4}\right)\right]\\ &=\sum_{k=0}^\infty\left[\frac3{(4k+1)(4k+4)}+\frac1{(4k+3)(4k+4)}\right] \end{align} $$ Which can be compared to $$ \left[\frac34+\frac3{16}\sum_{k=1}^\infty\frac1{k^2}\right] +\left[\frac1{12}+\frac1{16}\sum_{k=1}^\infty\frac1{k^2}\right] =\frac56+\frac14\sum_{k=1}^\infty\frac1{k^2} $$ which converges by the $p$-test.

One Approach to Evaluation

If curious about the actual sum, it can be computed, using $(11)$ from this answer, as $$ \begin{align} &\frac14\sum_{k=1}^\infty\left[-\left(\frac1k-\frac1{k-\frac34}\right)-\left(\frac1k-\frac1{k-\frac14}\right)\right]\\ &=\frac14\left[-H_{-3/4}-H_{-1/4}\right]\\ &=\frac14\left[-(-\pi/2-3\log(2))-(\pi/2-3\log(2))\right]\\ &=\frac32\log(2) \end{align} $$

Another Approach to Evaluation

Using the fact that the alternating Harmonic Series converges to $\log(2)$, we get $$ \begin{align} \sum_{k=0}^\infty\left[\frac1{4k+1}+\frac1{4k+3}-\frac2{4k+4}\right] &=\sum_{k=0}^\infty\left[\frac1{4k+1}-\frac1{4k+2}+\frac1{4k+3}-\frac1{4k+4}\right]\\ &+\sum_{k=0}^\infty\left[\hphantom{\frac1{4k+1}}+\frac1{4k+2}\hphantom{\ \!+\frac1{4k+3}}-\frac1{4k+4}\right]\\ &=\log(2)+\frac12\log(2)\\[6pt] &=\frac32\log(2) \end{align} $$

Community
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robjohn
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The following argument shows that the series converges, and gives its sum:

$\hspace{.3 in}1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \cdots=\ln 2$ $\;\;\;$so

$\hspace{.27 in}\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+\frac{1}{10}-\frac{1}{12}+\cdots=\frac{1}{2}\ln 2$. $\;\;\;$Inserting zeros, we get

$\hspace{.26 in}0+\frac{1}{2}+0-\frac{1}{4}+0+\frac{1}{6}+0-\frac{1}{8}+0+\frac{1}{10}+\cdots=\frac{1}{2}\ln 2$.

Adding this to the original series gives

$\hspace{.26 in}1+0+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+0+\frac{1}{7}-\frac{1}{4}+\frac{1}{9}+0+\cdots=\frac{3}{2}\ln 2$, $\;\;$ so

$\hspace{.25 in} 1+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\frac{1}{9}+\frac{1}{11}-\frac{1}{6}+\cdots=\frac{3}{2}\ln 2$.

user84413
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Here is a general result:

Let $s$ be an infinite sum over a sequence $(a_n)_{n\in\mathbb N}$, where $a_n$ are a permutation of $\frac{(-1)^{k+1}}{k}$, but where the sub sequences $(a_{n_k})_{k\in\mathbb N},\space a_{n_k}=-\frac{1}{2k}$ and $(a_{m_k})_{k\in\mathbb N},\space a_{m_k}=\frac{1}{2k-1}$ exist with $n_k<n_{k+1}$ and $m_k<m_{k+1}$. So $$ s=1+\frac13+\frac15-\frac12+\frac17+\frac19+\frac1{11}-\frac{1}{4}+... $$ is such a sum, but $$ s=1-\frac14+\frac13-\frac12+... $$ isn't.

Now, for such a sequence, define $p_n$ as the number of plus signs up to $a_n$, $q_n$ as the number of minus signs up to $a_n$ and $l=\lim_{n\to\infty}\frac{p_n}{q_n}$.

We have: $$ \sum_{k=1}^{n} a_k=\sum_{k=1}^{p_n}\frac{1}{2k-1}-\sum_{k=1}^{q_n}\frac{1}{2k} $$ But also: $$ \sum_{k=1}^{p_n}\frac{1}{2k-1}=\sum_{k=1}^{2p_n}\frac{1}{k}-\sum_{k=1}^{p_n}\frac{1}{2k} $$ And therefore, if $H_n$ is the $n$-th harmonic number and using the well known result $H_n=\ln(n)+\gamma+\epsilon_n$ where $\lim_{n\to\infty}\epsilon_n=0$, we obtain: $$ \sum_{k=1}^{n} a_k=\sum_{k=1}^{2p_n}\frac{1}{k}-\sum_{k=1}^{p_n}\frac{1}{2k}-\sum_{k=1}^{q_n}\frac{1}{2k}=\\ \sum_{k=1}^{2p_n}\frac{1}{k}-\frac{1}{2}\sum_{k=1}^{p_n}\frac{1}{k}-\frac{1}{2}\sum_{k=1}^{q_n}\frac{1}{k}=\\ H_{2p_n}-\frac{1}{2}H_{p_n}-\frac{1}{2}H_{q_n}=\\ \ln{(2p_n)}+\gamma+\epsilon_{2p_n}-\frac{1}{2}\ln{(p_n)}-\frac{1}{2}\gamma-\frac{1}{2}\epsilon_{p_n}-\frac{1}{2}\ln{(q_n)}-\frac{1}{2}\gamma-\frac{1}{2}\epsilon_{q_n}=\\ \ln{(2)}+\frac{1}{2}\ln{\left(\frac{p_n}{q_n}\right)}+\epsilon_{2p_n}-\frac{1}{2}\epsilon_{p_n}-\frac{1}{2}\epsilon_{q_n} $$ And thus: $$ \lim_{n\to\infty}\sum_{k=1}^{n} a_k=\lim_{n\to\infty}\left(\ln{(2)}+\frac{1}{2}\ln{\left(\frac{p_n}{q_n}\right)}+\epsilon_{2p_n}-\frac{1}{2}\epsilon_{p_n}-\frac{1}{2}\epsilon_{q_n}\right)=\ln(2)+\frac{1}{2}\ln(l) $$ In the original sum, we have $l=2$ and therefore $s=\frac{3}{2}\ln(2)$

Redundant Aunt
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