# Infinite zeros in infinite series

The problem:

Given that

$$\frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} + \ldots$$

Prove

$$\frac{\pi}{3} = 1 + \frac{1}{5} - \frac{1}{7} - \frac{1}{11} + \frac{1}{13} + \frac{1}{17} + \ldots$$

My solution:

We know

\begin{align} \frac{\pi}{4} & = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} -\frac{1}{11} + \frac{1}{13} - \frac{1}{15} + \ldots \\ \\ \frac{\pi}{12} & = \frac{1}{3} - \frac{1}{9} + \frac{1}{15} - \frac{1}{21} + \frac{1}{27} -\frac{1}{33} + \frac{1}{39} - \frac{1}{45} + \ldots\\ \\ & = 0 + \frac{1}{3} + 0 + 0 - \frac{1}{9} + 0 + 0 + \frac{1}{15} + 0 + 0 - \frac{1}{21} \end{align}

now add them together:

\begin{align} \frac{\pi}{4} + \frac{\pi}{12} & = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - \frac{1}{11} + \frac{1}{13} - \frac{1}{15} + \ldots \\ \\ & + 0 + \frac{1}{3} + 0 + 0 - \frac{1}{9} + 0 + 0 + \frac{1}{15} + \ldots \\ \end{align}

and we will get:

\begin{align} \frac{\pi}{3} & = 1 + 0 + \frac{1}{5} - \frac{1}{7} + 0 -\frac{1}{11} + \frac{1}{13} + 0 + \ldots \\ & = 1 + \frac{1}{5} - \frac{1}{7} -\frac{1}{11} + \frac{1}{13} + \ldots \end{align}

My questions:

• I inserted/removed infinite zeros into/from the series, is that OK?

• My solution relies on the fact that $\Sigma a_n + \Sigma b_n = \Sigma (a_n + b_n)$ and $k \Sigma a_n = \Sigma k a_n$. Is this always true for convergent infinite series? If so, why is it? (yeah I know this is a stupid question, but since I'm adding infinite terms up, I'd better pay some attention.)

• Bouns question: Can I arbitrarily (arbitrariness isn't infinity, you know) insert/remove zeros into/from a convergent infinite series, without changing its convergence value?
• My apologies. Inserting zeros is definitely not the same as rearranging. Inserting zeros in the series as you have done will not change the value of the series. – sharding4 Jun 13 '17 at 2:22
• @sharding4 Thanks. This is quite intuitive :) But can we rigidly prove this statement, say, with epsilon-delta definition? – nalzok Jun 13 '17 at 2:28
• You could probably approach the sequence of partial sums as a Cauchy sequence. Then adding a finite, fixed number of zeros at fixed intervals is the same as inserting duplicate terms into a Cauchy sequence. You may need to adjust for the slower convergence, but the sequence will still be Cauchy. – sharding4 Jun 13 '17 at 2:35
• Your series result can also be viewed in terms of Dirichlet $L$ series and Euler Products. Leibniz' series for $\pi/4$ is also a familiar Dirichlet series evaluated at 1. This series also has the product representation $\prod_p (1-\chi(p)/p)^{-1}$ where $\chi(p)= \pm 1\equiv p\bmod 4$ – sharding4 Jun 13 '17 at 2:52
• I assume you mean something like "Prove $\displaystyle \frac\pi3 = \sum_{k=0}^\infty (-1)^k \left( \frac1{6k+1} + \frac1{6k+5} \right)$"? Because, at least to me, it's far from obvious at a glance how to continue the series based on just the first 6 terms. – Ilmari Karonen Jun 13 '17 at 10:47

1. Yes, that is ok.
2. Yes it is. The partial sum of $\sum_k (a_k+b_k)$ is the partial sum of $\sum_ka_k$ plus the partial sum of $\sum_k b_k.$ The result follows from the sum property for limits.
3. Yes. Adding zeros will only delay the inevitable convergence of the sequence of partial sums. Where you insert zeros, the sequence of partial sums will hold flat. For the $N$ you find in the proof of convergence of the original, simply replace with $N$ plus the number of zeros you inserted before the N-th and you'll have the same value that will be within $\epsilon$ of the number the sum converges to.

Both the given identities follow from the fact that $\frac{\pi}{\text{something}}$ is related with the integral over $(0,1)$ of a rational function. The first identity is a consequence of $$\frac{\pi}{4} = \arctan(1)=\int_{0}^{1}\frac{dx}{1+x^2} = \int_{0}^{1}\left(1-x^2+x^4-x^6+\ldots\right)\,dx$$ and for the second series we may perform the same manipulation in the opposite direction, leading to:

$$\begin{eqnarray*}\sum_{n\geq 0}\left(\frac{1}{12n+1}+\frac{1}{12n+5}-\frac{1}{12n+7}-\frac{1}{12n+11}\right)&=&\int_{0}^{1}(1+x^4-x^6-x^{10})\sum_{n\geq 0}x^{12n}\,dx\\&=&\int_{0}^{1}\frac{1+x^4}{1+x^6}\,dx\\&=&\int_{0}^{1}\left(\frac{1}{1+x^2}+\frac{x^2}{1+x^6}\right)\,dx\\(x\mapsto z^{1/3})\qquad &=&\frac{\pi}{4}+\frac{1}{3}\int_{0}^{1}\frac{dz}{1+z^2}=\color{red}{\frac{\pi}{3}}.\end{eqnarray*}$$

Indeed, we are just multipling the first series by $\frac{4}{3}$ :D

• Is it really correct to try to integrate termwise a series which diverges at one of the limits (despite convergence for all other points of the domain of integration)? – Ruslan Jun 13 '17 at 9:41
• @Ruslan: in the shown manipulation, we never deal with a divergent series. $\sum_{n\geq 0}\left(\frac{1}{12n+1}+\ldots\right)$ is an absolutely convergent series and $\frac{1+x^4}{1+x^6}$ is a continuous and bounded function over $[0,1]$, so there is no issue. – Jack D'Aurizio Jun 13 '17 at 9:56
• I mean the first of the displayed equations, its rightmost side. The integrand diverges at $x=1$. – Ruslan Jun 13 '17 at 9:57
• About the arctangent, $\sum_{n=1}(-1)^n x^{2n}$ is not really divergent at $x=1$, since its partial sums are still bounded. Since the point $x=1$ has measure zero, we are allowed to perform the shown manipulation: if some $f(x)$ is continuous and bounded over $[0,1]$ and it is not really defined at $x=0$ or $x=1$, we may simply ignore the issue since $\int_{[0,1]}f(x)\,dx = \int_{(0,1)}f(x)\,dx$. – Jack D'Aurizio Jun 13 '17 at 9:58

It is a rewarding exercise to write, or read, detailed rigorous proofs of some simple, "obvious" results.

The definition of $x=\sum_{j=1}^{\infty}y_j$ is that $x=\lim_{n\to \infty}S_n$ where $S_n=\sum_{j=1}^ny_j.$ A useful way to state (or define) that the sequence $S=(S_n)_n$ converges to $x$ is that for any $r>0$ the set $$F(S, r)=\{n: S_n\not \in [-r+x,r+x]\}$$ is a finite set.

Let $x=\sum_{j=1}^{\infty}y_j.$ Insert some $0$'s into the sequence $(y_j)_j$ to produce a new sequence $(z_i)_i.$ For each $i$ we have either $z_i=0$ or $z_i=y_{g(i)}$ where $g(i)\leq i.$

Let $T=(T_m)_m$ where $T_m=\sum_{i=1}^mz_i.$

Let $i_0$ be the least (or any) $i$ such that $z_i=y_1.$

Then for every $m\geq i_0$ we have $T_m=\sum_{ \{g(i):i\leq m\}}y_{g(i)}= S_{f(m)}$ where $f(m)=\max \{g(i):i\leq m\}.$

Note that for any $n$ there exists $m_0$ such that $z_{m_0}=y_{n+1}$ so there exists $m_0$ such that $f(m_0)=n+1.$

Now for any $r>0$ we have $$F(T,r)=\{m:T_m\not \in [-r+x,r+x]\}\subset \{m\geq i_0:S_{f(m)} \not \in [-r+x.r+x]\}\cup \{m:m<i_0\}$$ $$=\{m\geq i_0: f(m)\in F(S,r)\}\cup \{m:m<i_0\}.$$ Observe that for any $n$ the set $\{m\geq i_0: f(m)=n\}$ is a finite set, because there exists $m_0$ such that $f(m_0)=n+1,$ and hence $m\geq m_0\implies f(m)>n.$ And recall that $F(S,r)$ is a finite set because the sequence $S$ converges to $x.$

So for any $r>0$ the set $F(T,r)$ is a subset of a union of a finite collection of finite sets: $$F(T,r)\subset (\cup_{n\in F(S,r)}\{m\geq i_0:f(m)=n\})\cup \{m:m<i_0\}.$$ Therefore $F(T,r)$ is finite for every $r>0$. Therefore $T$ converges to $x.$ Therefore $x=\sum_{i=1}^{\infty}z_i.$

• Some "obvious" results turn out to require long and difficult proofs. For example the Jordan Curve Theorem. – DanielWainfleet Jun 13 '17 at 16:43