Now I think the lhs can be rewritten as $$\sum _{n=1}^\infty \left( \dfrac {1} {2n-1}-\dfrac {1} {2\cdot 3^{n-1}}-\dfrac {1} {2^{n+1}}\right) =\dfrac {1} {2}\log 2$$

I guess one way to do this may be to start from RHS and try to bring up an expression which is same as the LHS. so using log series expansion $$\frac {1} {2}\log 2 = \log \sqrt {2} = \sum _{n=1}^\infty \dfrac {\left( -1\right) ^{n+1}} {n}\left( \sqrt {2}-1\right) ^{n}$$

Now Binomial expansion Theorem seems to be making a calling here. $$\sum _{n=1}^{\infty }\dfrac {\left( -1\right) ^{n+1}} {n}\left( \sqrt {2}-1\right) ^{n} = \sum _{n=1}^{\infty }\dfrac {\left( -1\right) ^{n+1}} {n}\sum _{k=0}^{k=n}C_{k}^{n} \sqrt {2}^{k}\left( -1\right) ^{n-k} $$

I am unsure how to proceed from here and even if i am going down the right path. Any help would be much appreciated.

  • $\begingroup$ Actually come to think of it we may be able to use limit along with sterling's formula to further solve this expression. possibly ? $\endgroup$ – Comic Book Guy Mar 10 '12 at 21:41
  • 2
    $\begingroup$ Actually, I think your left hand side should be interpreted as $\sum_{k=1}^{\infty} \left( \frac1{2k-1} - \frac1{4k-2} - \frac1{4k} \right)$. Edit: Additional evidence why you need to interpret it like this wolframalpha.com/input/… $\endgroup$ – user17762 Mar 10 '12 at 21:43
  • $\begingroup$ The left side cannot be as you write it, as the first term in the sum is divergent and the last two are convergent. The sum will therefore diverge (about as $\log n/2$) $\endgroup$ – Ross Millikan Mar 10 '12 at 21:50
  • 1
    $\begingroup$ This should solve your problem: en.wikipedia.org/wiki/Alternating_series $\endgroup$ – NoChance Mar 10 '12 at 22:01
  • 1
    $\begingroup$ @Hardy: Thanks for showing your prior work on the problem! Very interesting, and I'm tempted to ask my Calc I students to evaluate your final sum as a difficult/bonus question when they get to power series, but I'm sure I'd be publicly drawn and quartered for that. $\endgroup$ – JakeR Mar 10 '12 at 22:23

Rewrite $1-\dfrac {1} {2} -\dfrac {1} {4}+\dfrac {1} {3}-\dfrac {1} {6}-\dfrac {1} {8}+\dfrac {1} {5}-\cdots$ as

$$\left(1-\dfrac {1} {2}\right) -\dfrac {1} {4}+\left(\dfrac {1} {3}-\dfrac {1} {6}\right)-\dfrac {1} {8}+\left(\dfrac {1} {5} - \dfrac{1}{10}\right) - \dfrac{1}{12} \ldots$$


$$\frac{1}{2} - \frac{1}{4} + \frac{1}{6} - \frac{1}{8} + \frac{1}{10} - \frac{1}{12} + \cdots = \frac{1}{2}\left(\frac{1}{1} - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \cdots\right) = \frac{1}{2} \log(2).$$

Edit: More generally (without numbers), we write the left hand side as $\displaystyle\sum_{k=1}^\infty \frac{1}{2k-1} - \frac{1}{4k-2} - \frac{1}{4k} $ which is equal to $\displaystyle\sum_{k=1}^\infty \frac{1}{4k-2}- \frac{1}{4k} $ (since $\dfrac{1}{2k-1} - \dfrac{1}{4k-2} = \dfrac{1}{2k-1} - \left(\dfrac{1}{2k-1}\right)\dfrac{1}{2} = \left(\dfrac{1}{2k-1}\right)\dfrac{1}{2}$).

This is equal to $\left(\dfrac{1}{2}\right)\displaystyle\sum_{k=1}^\infty \frac{1}{2k-1}- \frac{1}{2k} $ $= \dfrac{1}{2}\displaystyle\sum_{k=1}^\infty \dfrac{(-1)^{k+1}}{k}$, or $\dfrac{\log(2)}{2}$ by examination of the power series of $\log$.


The specific series of the question has been addressed by other answers. I am not including proofs, but I figure some additional information may be useful.

I recently learned from Marion Scheepers of some interesting results of which this is a particular case. The question of rearranging conditional series was carefully studied in the XIX and early XX centuries, and there seems to be a resurgence of interest in recent years. All I knew until a few months ago was Riemann's rearrangement theorem.

Here are some particular cases of what follows: We have (from Leibniz test) that the following three series converge conditionally: $$ \sum_{n=1}^\infty\frac{(-1)^{n-1}}n,\quad\sum_{n=1}^\infty\frac{(-1)^{n-1}}{\sqrt n},\quad\sum_{n=2}^\infty\frac{(-1)^n}{n\ln n}. $$ However, the rearrangement $$ 1-\frac12-\frac14+\frac13-\frac16-\frac18\dots $$ converges to $1/2$ of the original series; the rearrangement $$ \frac1{2\ln2}-\frac1{3\ln3}-\frac1{5\ln5}+\frac1{4\ln4}-\frac1{6\ln6}-\frac1{8\ln8}\dots $$ converges to exactly the same value as the original series; and the rearrangement $$ 1-\frac1{\sqrt2}-\frac1{\sqrt4}+\frac1{\sqrt3}-\frac1{\sqrt6}-\frac1{\sqrt8}\dots $$ diverges. One would imagine that either this is all very chaotic, or else there is some underlying theory worthy of study.

The following is Martin Ohm's theorem from 1839:

For $p$ and $q$ positive integers rearrange the sequence $$\left(\frac{(−1)^{n-1}} n\right)_{n\ge 1} $$ by taking the first $p$ positive terms, then the first $q$ negative terms, then the next $p$ positive terms, then the next $q$ negative terms, and so on. The rearranged series converges to $$ \ln(2) + \frac12 \ln\left( \frac pq \right). $$

This was generalized by Oscar Schlömilch in 1873 as follows:

Let's say that a sequence $c_1,c_2,\dots$ of real numbers is signwise monotonic iff the sequence $|c_1|,|c_2|,\dots$ of absolute values is monotone. Write $a_1,a_2,\dots$ for the subsequence of positive terms, and $-b_1,-b_2,\dots$ for the subsequence of negative terms.

Let $(c_n)_{n\ge1}$ be a signwise monotonic and suppose that $\sum_{n=1}^\infty c_n$ is conditionally convergent. For $p$ and $q$ positive integers, rearrange the sequence by taking the first $p$ positive terms, then the first $q$ negative terms, and so on. The rearranged series converges to $$ \sum_{n=1}^\infty c_n + g\ln\left(\frac pq\right), $$ where $g$ is the limit $\lim_{n\to\infty}n a_n$, if it exists.

Soon after, Alfred Pringsheim obtained some general results. In 1883 he proved three theorems (one of which was discovered defective as a consequence of Marion's work). Here is a quick summary of some of Marion's results, that extend Pringsheim's work:

For signwise monotone sequences $c_1,c_2,\dots$, given a subset $A\subseteq{\mathbb N}$, let $f_A(n)=a_j$ if $n$ is the $j$th element of $A$, and $f_A(n)=-b_j$ if $n$ is the $j$th element of ${\mathbb N}\setminus A$.

Write $d(A)$ for $\displaystyle \lim_{n\to\infty}\frac {|\{k\in A\mid k\le n\}|}n$, if it exists.

Let $0<x<1$. Then the following are equivalent for a signwise monotonic sequence converging to 0:

  1. $\lim_{n\to\infty}n a_n=\infty$, and there is some $A\subseteq{\mathbb N}$ such that $d(A)=x$ and $\sum_{n=1}^\infty f_A(n)$ converges.

  2. $d(B)=x$ for each set $B$ such that $d(B)$ exists and $\sum f_B$ converges.

A curious "anti-rearrangement" result holds for this density notion:

If there is more than one $x$ such that $0<x<1$ and for some $A$, $d(A)=x$ and $\sum_n f_A(n)$ converges, then for any $B,C$, if $d(B)$ exists, $d(B)=d(C)$, and $\sum_n f_B(n)$ converges, then $\sum_n f_C(n)$ also converges, and $$ \sum_n f_B(n)=\sum_n f_C(n). $$

In this case, call $\Phi(x)$ the value $\sum_n f_B(n)$ for any $B$ with $d(B)=x$ (if this sum converges for some such $B$, otherwise, $\Phi(x)$ is undefined).

Then we have:

Suppose we have a signwise monotone sequence converging to 0. Let $t\in{\mathbb R}$ be a fixed real. The following are equivalent:

  1. $\lim_n n\cdot a_n=0$, and there is some $x$ with $0<x<1$ and there is some $A$ such that $d(A)=x$ and $\sum_n f_A(n)=t$.

  2. For each $B$, if $d(B)$ exists and $0<d(B)<1$, then $\sum_n f_B(n)=t$.

  3. $\Phi(x)=t$ for all $x\in(0,1)$.

  • $\begingroup$ wow , thank you so much for this insightfull and compilation. I recently considered a variant of harmonic series where p+q negative terms allways followed p positive terms and was wondering about the series with similar setup as the one u described, did n't know there was a theorem about that. I am still reading $\endgroup$ – Comic Book Guy Mar 11 '12 at 0:44
  • $\begingroup$ I had only heard of Alfred Pringsheim's theorem of double series convergence. If you unfamiliar with that result i could post that here. I had not idea he proved other relevant results. $\endgroup$ – Comic Book Guy Mar 11 '12 at 1:00
  • $\begingroup$ @Hardy: Sure, it would be nice to see the result you mentioned. $\endgroup$ – Andrés E. Caicedo Mar 11 '12 at 3:22
  • $\begingroup$ Mate it's my birthday today and the wife expects some time and to do this one right would require some work, so would post it tomz Stay tuned. $\endgroup$ – Comic Book Guy Mar 11 '12 at 5:58

I think your left hand side should be interpreted as $$\sum_{k=1}^{\infty} \left( \frac1{2k-1} - \frac1{4k-2} - \frac1{4k} \right).$$ Then note that $$a_k = \frac1{2k-1} - \frac1{4k-2} - \frac1{4k} = \frac1{2(2k-1)} - \frac1{4k} = \frac12 \left( \frac1{2k-1} - \frac1{2k} \right).$$ Do you see it now why it is $\frac12 \log(2)$?

  • $\begingroup$ Thanks for your help buddy. I could only pick 1 answer, but your answer was educational as well, and yes i get why the series equates to that. $\endgroup$ – Comic Book Guy Mar 10 '12 at 22:08
  • $\begingroup$ +1 because it doesn't give too much away (making the reader figure out the exact connection to log(2), which is in their benefit) while giving a general strategy for the future to analyze seemingly-unfamiliar sums by breaking it into familiar ones. Nice answer. $\endgroup$ – JakeR Mar 10 '12 at 22:18

Double Series Let $u_{m,n}$ be a number determinate for all positive integral values of m and n; consider the array $$u_{1,1},u_{1,2},u_{1,3},\ldots$$



Let the sum of the terms inside the rectangle, formed by the first $m$ rows of the first $n$ columns of this array of terms, be denoted by $S_{m,n}$.

If a number $S$ exists such that, given any arbitrary positive number $\epsilon$, it is possible to find integers $m$ and $n$ such that $$\left| S_{\mu ,v} -S\right| < \epsilon $$ whenever both $\mu > m$ and $v>n$, we say that the double series of which the general element is $u_{\mu ,v}$ converges to the sum $S$,and we write $$\lim _{\mu\rightarrow \infty ,v\rightarrow \infty }S_{\mu,v}=S$$. This definition is practically due to Cauchy.

If the double series, of which the general element is $\left|u_{\mu,v}\right|$, is convergent, we say that the given double series is absolutely convergent.

Since $$u_{\mu,v} = \left(S_{\mu,v} - S_{\mu,v-1}\right)\left(S_{\mu - 1,v} - S_{\mu -1,v -1}\right)$$, it is easily seen that, if the double series is convergent, then $$\lim _{\mu\rightarrow \infty ,v\rightarrow \infty }u_{\mu ,v}=0.$$

Stolz'necessary and sufficient condition for convergence (But first proven by Pringsheim). A condition for convergence which is obviously necessary is that, given $\epsilon$, we can find m and n such that $$\left|S_{\mu + \rho,v+\sigma} - S_{\mu,v-1}\right| < \epsilon$$ whenever $\mu > m$ and $v>n$ and $\rho,\sigma$ may take any of the values 0,1,2,....

The condition is also sufficient; for, suppose it satisfied; then, when $\mu>m +n$,$\left|S_{\mu + \rho,v+\rho} - S_{\mu,\mu}\right| < \epsilon$. Therefore, $S_{\mu,\mu}$ has a limit $S$; then, making $\rho$ and $\sigma$ tend to infinity in such a way that $\mu + \rho = v + \sigma$, we see that $\left|S - S_{\mu,v}\right|\leq \epsilon$ whenever $\mu > m$ and $v >n$; that is to say, the double series converges.

An absolutely convergent double series is convergent. For if the double series converges absolutely and if $t_{m,n}$ be the sum of $m$ rows and $n$ columns of the series of moduli, then,given $\epsilon$, we can find $\mu$ such that, when $p>m>\mu$ and $q>n>\mu$,$t_{p,q}-t_{m,n}<\epsilon$, but $\left| S_{p,q}-S_{m,n}\right| \leq t_{p,q}-t_{m,n}$ and so $\left| S_{p,q}-S_{m,n}\right| < \epsilon $ when $p>m>\mu, q>n>\mu$; and this is the condition that the double series should converge.

Methods of Summing double Series Let us suppose that $\sum _{v=1}^{\infty }U_{\mu,v}$ converges to the sum $S_{\mu}$. Then $\sum _{\mu=1}^{\infty }S_{\mu}$ is called the sum by rows of the double series; that is to say, the sum by rows is $\sum _{\mu =1}^{\infty }\left( \sum _{v=1}^{\infty }u_{\mu },v\right) $. Similarly, the sum by columns is defined as $\sum _{v=1}^{\infty }\left( \sum _{\mu =1}^{\infty }u_{\mu },v\right) $. That these two sums are not necessarily the same is shown by the example $S_{\mu ,v}=\dfrac {\mu - v } {\mu+v}$, in which the sum by rows is -1, the sum by columns is +1; and S does not exist.

Pringsheim's Theorem: If $S$ exists and the sums by rows and sum by columns exist; then each of these sums is equal to S. For since $S$ exists, then we can find $m$ such that $$\left| S_{\mu ,v} -S\right| < \epsilon $$ whenever both $\mu > m$ and $v>m$. And therefore, since $\lim _{v\rightarrow \infty }S_{\mu ,v}$ exists, $\left| \left( \lim _{v\rightarrow \infty }S_{\mu ,v}\right) -S\right| \leq \epsilon$; that is to say, $\left| \sum _{p=1}^{\mu }S_{p}-S\right| \leq \epsilon $ when $\mu>m$ and so the sum by rows converges to $S$. In a like manner the sum of columns converge to $S$.

  • $\begingroup$ It got a bit long so instead of posting it as comments i think it would be better suited as an answer submission. $\endgroup$ – Comic Book Guy Mar 11 '12 at 21:20

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.