# Problem when proving the series $1+\frac 13-\frac 12+\frac 15+\frac17-\frac 14 +\frac 19+\frac 1{11}-\frac 16+...$ converges to $\frac 32\log 2.$

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 ?

• Your "$\lim S=0$" is exactly the mistake. Commented Jun 16, 2023 at 5:59
• @metamorphy Why? Commented Jun 16, 2023 at 6:00
• The third term in your five-times-written-out series is wrong. Commented Jun 16, 2023 at 6:01
• @JohnBentin It will be helpful, if you specifically mention which term. It looks confusing. Are you talking about the term $-\frac 12$ ? Commented Jun 16, 2023 at 6:02
• 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. Commented Jun 16, 2023 at 7:27

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 $$m_l for every $$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 $$2$$.

– Koro
Commented Jun 16, 2023 at 11:31
• 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, $\sum_na_n=\sum_nb_n$. Commented Jun 18, 2023 at 2:47
• 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.
– Koro
Commented Jun 18, 2023 at 4:32
• @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.
– Koro
Commented Jun 18, 2023 at 4:33
• @Mittens: I have added the details in the post.
– 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.

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