Prove that the series $$1+\frac 13-\frac 12+\frac 15+\frac17-\frac 14 +\frac 19+\frac 1{11}-\frac 16+\cdots$$ converges to $\frac 32\log 2.$

I tried solving the problem as follows:

The series given is $$1+\frac 13-\frac 12+\frac 15+\frac17-\frac 14 +\frac 19+\frac 1{11}-\frac 16+\cdots.$$

We write the given series as $\sum u_n$ where $\{u_n\}$ is a sequence defined by $u_n=\frac{1}{4n-3}+\frac{1}{4n-1}-\frac{1}{2n},\forall n\in\Bbb N.$

We consider the partial sum of the given series $t_n=u_1+u_2+u_3+...+u_n$( consisting of $n$ terms).

Then, $$t_{3n}=u_1+u_2+\cdots +u_{3n}=(1+\frac 13-\frac 12)+(\frac 15+\frac17-\frac 14) +(\frac 19+\frac 1{11}-\frac 16)+\cdots+(\frac{1}{12n-3}+\frac{1}{12n-1}-\frac{1}{6n})=(1+\frac 13+\frac 15+\frac 17+...+\frac{1}{12n-1})-(\frac 12+\frac 14+\frac 16+\cdots +\frac 1{6n})\tag 1.$$

(Till now, we have grouped all the odd terms (positive terms) together and the negative terms( even terms) together in the partial sum, above.) Next, we add and subtract $\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{12n}$ in the right hand side of the above equality $(1)$ and rearrange, to get the partial sum $t_{3n}$ of the series as,

$$t_{3n}=(1+\frac 13+\frac 15+\frac 17+...+\frac{1}{12n-1})-(\frac 12+\frac 14+\frac 16+\cdots +\frac 1{6n})=(1+\frac 12+\frac 13+\cdots +\frac{1}{12n-1})-(\frac 12+\frac 14+\frac 16+\cdots +\frac {1}{12n})-(\frac 12+\frac 14+\frac 16+\cdots +\frac 1{6n})=(1+\frac 12+\frac 13+\cdots +\frac{1}{12n-1})-2(\frac 12+\frac 14+\frac 16+\cdots +\frac {1}{6n})-(\frac1{6n+2}+\frac{1}{6n+4}+\cdots+\frac{1}{12n})=(1+\frac 12+\frac 13+\cdots +\frac{1}{12n-1})-( 1+\frac 12+\frac 13+\cdots +\frac {1}{3n})-(\frac1{6n+2}+\frac{1}{6n+4}+\cdots+\frac{1}{12n})\tag 2$$

We know that, $\lim(1+\frac 12+\frac 13+\cdots+ \frac 1n-\log n)=\gamma.$ Let, $1+\frac 12+\frac 13+\cdots+ \frac 1n-\log n=\gamma_n\tag 3$ then, $\lim\gamma_n=\gamma.$

Using $(3),$ the equality $(2)$ can be written as, $$t_{3n}=(1+\frac 12+\frac 13+\cdots +\frac{1}{12n-1})-( 1+\frac 12+\frac 13+\cdots +\frac {1}{3n})-(\frac1{6n+2}+\frac{1}{6n+4}+\cdots+\frac{1}{12n})=\log(12n-1)+\gamma_{12n-1}-\log(3n)-\gamma_{3n}-S,$$ where $S=\frac1{6n+2}+\frac{1}{6n+4}+\cdots+\frac{1}{12n}.$

We also, note that, $\lim S=0.$

We now proceed to evaluate the limit of $t_{3n}=\log(12n-1)+\gamma_{12n-1}-\log(3n)-\gamma_{3n}-S=\log(4-\frac{1}{3n})+\gamma_{12n-1}-\gamma_{3n}+S.$ We observe,

$$\lim t_{3n}=\log 4=2\log 2.$$

As, $t_{3n+1}=t_{3n}+u_{3n+1}$ and $t_{3n+2}=t_{3n+1}+u_{3n+2},$ we have, $$\lim t_{3n+1}=\lim t_{3n+2}=\lim t_{3n}=2\log 2.$$

So, $\lim t_n=2\log 2.$ Hence, the series $$1+\frac 13-\frac 12+\frac 15+\frac17-\frac 14 +\frac 19+\frac 1{11}-\frac 16+\cdots$$ converges to $2\log 2.$

However, as it turns out, this gives the value $2\log 2$ contrary to what was required to be established, i.e $\frac 32\log 2.$ The problem, is precisely, I find no mistake in my solution. I want to know where did things go wrong. Specifically, where is the mistake in this solution?

A New Issue

As pointed out, in the comment, by @metamorphy, it seems that the evaluation of the limit .i.e $\lim S=0$ is incorrect. It should be, as follows:

$$S=\frac1{6n+2}+\frac{1}{6n+4}+\cdots+\frac{1}{12n}\implies 2S=\frac1{3n+1}+\frac1{3n+2}+\dots+\frac1{6n}\\=\left(1+\frac12+\dots+\frac1{6n}\right)-\left(1+\frac12+\dots+\frac1{3n}\right)\\=\big(\gamma_{6n}+\log(6n)\big)-\big(\gamma_{3n}+\log(3n)\big).$$ This means, $\lim 2S=\log 2,$ or $\lim S=\frac 12 \log 2.$

But here's the problem, what went wrong with this reasoning, i.e the reason by which, I concluded $\lim S$ is $0$ which is:

$$S=\frac1{6n+2}+\frac{1}{6n+4}+\cdots+\frac{1}{12n},$$ so, limit of each of the terms $\lim\frac{1}{6n+r}=0$, such that $r\in\{2,4,...,6n\}.$ This, means $\lim S=0.$

I want to know the mistake in this second limit evaluation explicitly ?

  • $\begingroup$ Your "$\lim S=0$" is exactly the mistake. $\endgroup$
    – metamorphy
    Commented Jun 16, 2023 at 5:59
  • $\begingroup$ @metamorphy Why? $\endgroup$ Commented Jun 16, 2023 at 6:00
  • $\begingroup$ The third term in your five-times-written-out series is wrong. $\endgroup$ Commented Jun 16, 2023 at 6:01
  • $\begingroup$ @JohnBentin It will be helpful, if you specifically mention which term. It looks confusing. Are you talking about the term $-\frac 12$ ? $\endgroup$ Commented Jun 16, 2023 at 6:02
  • 1
    $\begingroup$ I'm just observing; I'm not implying anything. Your questions seem fine and in the spirit of the site. I also think what you're doing by putting further questions into the body of this question to avoid make new questions is a good idea. $\endgroup$ Commented Jun 16, 2023 at 7:27

4 Answers 4


There is no issue. You do $\lim_n (x_n+y_n)= \lim x_n +\lim y_n$ when there are finitely many (fixed) summands. Clearly, $1= 1/n +1/n +1/n+...+1/n$ but $\lim (1/n+1/n+...+1/n)\ne \color{red}{\lim 1/n +\lim 1/n+...+\lim 1/n= 0}$. So you made an error in computing $\lim S$.

Here is a generalisation: taking $a$ 'odd' terms followed by $b$ 'even' negative terms.

Specifically, consider the series $$1+1/3+...+\frac 1{2a-1}-\color{blue}{1/2-1/4-...-\frac 1{2b}}+\frac 1{2a+1}+...+\frac 1{4a-1}-\color{blue}{\frac 1{2b+2}-...-\frac 1{4b}}+...$$ $\tag A$

Result $1$: The above series has sum $=\log \left(2\sqrt \frac ab\right)$.

Pf: Consider the $(a+b)n$-th partial sum

$$S_{(a+b)n}=1+1/3+...+\frac 1{2a-1}-\left(1/2+1/4+...+\frac 1{2b}\right)+\cdots+\frac 1{2(n-1)a+1}+\cdots\\+\frac 1{2na-1}-\left(\frac 1{2b(n-1)+2}+\cdots+\frac 1{2nb}\right)$$

$S_{(a+b)n}= H_{2na}- \left(\frac 12+\frac 14+\cdots+\frac 1{2na}\right)-\frac 12 H_{nb}= H_{2na}-\frac 12 H_{na}-\frac 12 H_{nb}$

$S_{(a+b)n}= \log 2na - \frac{\log na +\log nb}{2}+o(1)\to \log\left(2\sqrt{\frac ab}\right)$ as $n\to \infty$.

(B) It can be shown that for every $m\in \{1,2,...,a+b-1\}, S_{(a+b)n+m}\to \log \left(2\sqrt \frac ab\right)$.

It follows that the series has sum $=\log \left(2\sqrt \frac ab\right). \square$

To compute the sum of the series in your post: put $a=2, b=1$.

(B) To see why it answers the question, note the following result:

Result 2: Let $\{m_l\}, \{n_k\}$ be exhausting, disjoint subsets of $\mathbb N$ (by exhausting it is meant that the union of the indices is $\mathbb N$) such that $m_l<m_{l+1}, n_k<n_{k+1}$ for every $l,k\in \mathbb N$. Then the sequence $(x_n), x_n\in \mathbb R$ converges iff the subsequences $(x_{m_l}), (x_{n_k})$ converge to the same limit. (Let's call such subsequences 'complementary subsequences')

Let's see how this result alongwith (B) completes the proof of Result $1$. Suppose Result $2$ is true, then we can extend it any finite collection of complementary subsequences. Let's use this on the sequence of partial sums $(S_n)$ of the series in $(A)$. The subsequences $(S_{(a+b)n+m})$ in (B) are complementary and have the same sum viz. $\log \left(\sqrt{\frac ab} \right)$ so we are done.

Pf. of Result $2$: $(\Rightarrow)$ Every subsequence of a convergent sequence converges to the same limit so this part is proved.

$(\Leftarrow)$ Suppose on the contrary that the sequence $(x_p)$ doesn't converge. Suppose the complementary subsequences converge to $L$. It follows that there exists some $\epsilon_0$ and a subsequence $(x_{p_t})_{t=1}^\infty$ such that $|x_{p_t}-L|\ge \epsilon_0$ for all $t\in \mathbb N$.

By the convergence of complementary subsequences, there exists $N\in \mathbb N$ such that for all $m_l, n_k\gt N$, $|x_{m_l}-L|<\epsilon_0, |x_{n_k}-L|<\epsilon_0$. Since the indices $\{m_l\}, \{n_k\}$ are exhausting, assume wlog that there exists some $m_k\gt N$ such that $x_{m_k}= x_{p_T}$ for some $T$, so we must have $|x_{p_T}-L|=|x_{m_k}-L|<\epsilon_0$, which is a contradiction.

This proves Result $2$.

  • $\begingroup$ @ThomasFinley: See my comment down your post. This answers your question and gives a generalisation also, thus answering your latest post as well :-) $\endgroup$
    – Koro
    Commented Jun 16, 2023 at 11:31
  • 1
    $\begingroup$ One still needs to justify that if by introducing parentheses to a series $\sum_na_n$ convergence to a number $s$ occurs, then the original series converges and that the summation is also $s$. To be more precise, if for some monotone increasing function $f:\mathbb{N}\rightarrow\mathbb{N}$, one defines $b_1=a_1+\ldots + a_{p(1)}$, and $b_n=a_{p(n)+1}+\ldots + a_{p(n+1)}$ for $n\geq2$, under some conditions, $\sum_na_n$ converges iff $\sum_nb_n$ converges and in such a case, $\sum_na_n=\sum_nb_n$. $\endgroup$
    – Mittens
    Commented Jun 18, 2023 at 2:47
  • $\begingroup$ 1) the purpose of parentheses in the original series was to show pattern of the terms in the series but it wasn't mentioned and is not even a common practice so I removed them and colorcoded (in blue) the terms instead. 2) Rearrangement here is not an issue: a) Lemma: if $\{n_l\}$ and $\{m_k\}$ are exhausting disjoint indices (i.e. $n_l < n_{l+1}, m_k< m_{k+1}, \mathbb N= \cup_l\cup_k \{n_l\}\cup \{m_k\} $, then the sequence $(x_n)$ converges iff the subsequences $(x_{n_l}), (x_{m_k})$ converge to the same limit. $\endgroup$
    – Koro
    Commented Jun 18, 2023 at 4:32
  • $\begingroup$ @Mittens (Contd.) b) note that the indices in $S_{(a+b)n+m}, m= 1,2,...,a+b-1$ are disjoint, exhaust $\mathbb N$ and for every $m$, $S_{(a+b)n+m}$ converges to $\log \left(2\sqrt{\frac ab}\right)$so by the lemma in a), the sequence of partial sums $(S_n)$ converges hence we are done. $\endgroup$
    – Koro
    Commented Jun 18, 2023 at 4:33
  • $\begingroup$ @Mittens: I have added the details in the post. $\endgroup$
    – Koro
    Commented Jun 18, 2023 at 5:06

This is a problem on introducing parenthesis on a series. This technique is very common and is justified by the following result:

Lemma I: Suppose $p:\mathbb{N}\rightarrow\mathbb{N}$ is a strictly monotone increasing function. Let series $\sum_na_n$ and $\sum_nb_n$ be series related as follows \begin{align} b_1&=a_1+\ldots+a_{p(1)}\\ b_n&=a_{p(n)+1}+\ldots a_{p(n+1)}, \qquad n\geq2 \end{align} If $M:=\sup_n(p(n+1)-p(n))<\infty$ and $a_n\xrightarrow{n\rightarrow\infty}0$, the series $\sum_na_n$ converges iff the series $\sum_nb_n$ converges. In such case, the sum is the same.

The series in the OP $$\sum_na(n)=1+\frac13-\frac12+\frac15+\frac17-\frac14+\ldots$$ has its $n$-th term defined as \begin{align} a(3n)&=-\frac{1}{2n}\\ a(3n-1)&=\frac{1}{2(2n)-1}\\ a(3n-2)&=\frac{1}{2(2n-1)-1} \end{align} for $n\in\mathbb{N}$.

Define $p(n)=3n$, $n\in\mathbb{N}$ and consider adding parenthesis of length 3 to the series $\sum_na_n$, that is \begin{align} b_1&=a(1)+a(2)+a(3)=1+\frac13-\frac12\\ b_n&=a(3n+1)+a(3n+2)+a(3n+3)=\frac{1}{4n+1}+\frac{1}{4n+3} -\frac{1}{2(n+1)} \end{align}

Clearly $a(n)\xrightarrow{n\rightarrow\infty}0$. On the other hand, $t_n:=\sum^n_{n=1}b_n$ can be expressed as $$t_n=\sum^{2n+1}_{k=1}\frac{1}{2k+1}-\frac12\sum^{n+1}_{k=1}\frac{1}{k}$$

Now we exploit the scaling properties of the harmonic series $\sum_n\frac1n$. Let $H_n:=\sum^n_{k=1}\frac{1}{k}$. It is well known that $$H_n=\log n+\gamma +O(\frac1n)$$ where $\gamma$ is some constant. Then \begin{align} t_n&=H_{4n+3}-\frac12H_{2n+1}-\frac12 H_{n+1}\\ &=H_{4n+3}-H_{n+1} -\frac12(H_{2n+1}-H_{n+1})\\ &=\log\big(\frac{4n+3}{n+1}\big)-\frac12\log\big(\frac{2n+1}{n+1}\big)+O\big(\frac1n\big) \end{align} Passing to the limit, we obtain that $$t_n\xrightarrow{n\rightarrow\infty}\log 4-\frac12\log 2=\frac32\log2$$ By the Lemma above, we obtain that $$\sum_na_n=\sum_nb_n=\frac32\log2$$

No that the convergence if the series $\sum_na_n$ has been established, it is also true that any other system of parenthesis (with bounded or unbounded length) would produce the same result:

Lemma II: Is a series $\sum_na_n$ converges to $s$, then every series $\sum_nb_n$ obtained from $\sum_na_n$ by inserting parentheses converges to $s$.

The Lemmas above are not difficult to prove, they are discussed in many Calculus and Analysis textbooks. See for example Apostol, T., Mathematical Analysis, 2nd Edition, Addison-Wesley, 1974, pp. 187.


Concerning the Question

One small problem (which washes out since $\frac1{12n}\to0$) is that, in $(2)$, there is $$ \scriptsize\left[\left(1+\frac12+\frac13+\cdots+\frac1{12n-1}\right)-\left(\frac12+\frac14+\frac16+\cdots+\color{#C00}{\frac1{12n}}\right)\right]-\left(\frac 12+\frac 14+\frac16+\cdots+\frac1{6n}\right) $$ The bracketed piece is supposed to be the sum of the reciprocals of the odd numbers up to $12n-1$. However, there is a $-\frac1{12n}$ that is not cancelled. What should be there is $$ \scriptsize\left(1+\frac12+\frac13+\cdots+\color{#C00}{\frac1{12n}}\right)-\left(\frac12+\frac14+\frac16+\cdots+\color{#C00}{\frac1{12n}}\right)-\left(\frac 12+\frac 14+\frac16+\cdots+\frac1{6n}\right) $$ So that the $-\frac1{12n}$ is cancelled.

At this point, the sum is $$ \begin{align} &(\log(12n)+\gamma)-\frac12(\log(6n)+\gamma)-\frac12(\log(3n)+\gamma)+O\!\left(\frac1n\right)\\ &=\frac32\log(2)+O\!\left(\frac1n\right) \end{align} $$ which leads to the proper sum (as I said earlier, this washes out). In fact, the rest of $(2)$ is okay. Therefore, any problem must lie further on.

The New Issue

The mistake is saying that $$ \lim_{n\to\infty}\left(\frac1{6n+2}+\frac1{6n+4}+\cdots+\frac1{12n}\right)=0 $$ This is actually $$ \begin{align} &\frac12\left(\left(1+\frac12+\frac13+\cdots+\frac1{6n}\right)-\left(1+\frac12+\frac13+\cdots+\frac1{3n}\right)\right)\\ &=\frac12((\log(6n)+\gamma)-(\log(3n)+\gamma))+O\!\left(\frac1n\right)\\ &=\frac12\log(2)+O\!\left(\frac1n\right) \end{align} $$ and there is the extra $\frac12\log(2)$.

For a fixed $k$, if $\lim\limits_{n\to\infty}a_n=0$, then $$ \lim\limits_{n\to\infty}(a_n+a_{n+1}+a_{n+2}+\dots+a_{n+k})=0 $$ In the case at hand, $k$ varies with $n$.

A Similar Approach

We will use the approximation $$ \sum_{k=1}^n\frac1k=\log(n)+\gamma+O\!\left(\frac1n\right)\tag1 $$ We will group the terms into groups of three as in the question, which is okay since the terms tend to $0$. $$ \begin{align} \sum_{k=1}^n\left(\frac1{4k-3}+\frac1{4k-1}-\frac1{2k}\right) &=\overbrace{\color{#C00}{\sum_{k=1}^{4n}\frac1k}-\color{#090}{\sum_{k=1}^{2n}\frac1{2k}}}^{\text{odd terms to $4n-1$}}-\overbrace{\ \color{#B80}{\sum_{k=1}^{\vphantom{4}n}\frac1{2k}}\ }^{\substack{\text{even terms}\\\text{to $2n$}}}\tag{2a}\\ &=\color{#C00}{(\log(4n)+\gamma)}-\color{#090}{\frac12(\log(2n)+\gamma)}\\ &-\color{#B80}{\frac12(\log(n)+\gamma)}+O\!\left(\frac1n\right)\tag{2b}\\ &=\log(4)-\frac12\log(2)+O\!\left(\frac1n\right)\tag{2c}\\ &=\frac32\log(2)+O\!\left(\frac1n\right)\tag{2d} \end{align} $$ Explanation:
$\text{(2a):}$ regroup the terms (allowed because the sum is finite)
$\text{(2b):}$ apply $(1)$, grouping the error terms
$\text{(2c):}$ collect and cancel
$\text{(2d):}$ simplify using $\log(4)=2\log(2)$

Taking the limit of $(2)$ as $n\to\infty$ gives $$ \sum_{k=1}^\infty\left(\frac1{4k-3}+\frac1{4k-1}-\frac1{2k}\right) =\frac32\log(2)\tag3 $$


Let us define two sequences:

$$ a_n:=\frac{(-1)^{n+1}}{n},\qquad b_n:=\begin{cases}0 &\text{if $n\equiv 1\pmod 2$,}\\[10pt]\dfrac{(-1)^{n/2+1}}{n}&\text{if $n\equiv 0\pmod 2$.}\end{cases} $$

Then \begin{align*} \sum_{n=1}^{\infty}a_n&=1-\frac 12+\frac 13-\frac 14+\frac 15-\frac 16+\frac 17-\frac 18+\ldots=\log 2\\[10pt] \sum_{n=1}^{\infty}b_n&=0+\frac 12+0-\frac 14+0+\frac 16+0-\frac 18+\ldots=\frac 12\sum_{n=1}^{\infty}a_n=\frac 12\log 2\\ \end{align*}

Adding these squences we get $$ a_n+b_n= \begin{cases} \dfrac 1n &\text{if $n\equiv 1\pmod 4$,}\\[10pt] 0 &\text{if $n\equiv 2\pmod 4$,}\\[10pt] \dfrac 1n &\text{if $n\equiv 3\pmod 4$,}\\[10pt] -\dfrac 2n &\text{if $n\equiv 0\pmod 4$.} \end{cases} $$

The corresponding series is the series which is discussed.

$$ \sum_{n=1}^{\infty}(a_n+b_n)=1+0+\frac 13-\frac 12+\frac 15+0+\frac 17-\frac 14+\frac 19+0+\frac 1{11}-\frac 16+\ldots=\frac 32\log 2. $$


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