# Riemann Series Theorem

**Riemann series theorem** is named after a great German mathematician **Bernhard Riemann** who contributed a lot to mathematics in the fields of analytical number theory and calculus. In 1859, he gave a paper on prime counting function, which is considered as one of the most influential papers in number theory. The **Riemann series theorem** tells us that if an infinite series is conditionally convergent, then given any value $r$, the terms can be permuted so that the series converges to $r$. This is a surprising result since it is obviously true that when we have finitely many terms, permuting the terms doesn't cause the same to change, and it is not clear why having infinitely many terms could affect the sum.

## Definitions

Consider a sequence $x_{1}, x_{2}, x_{3}, x_{4}, \ldots,$ and let $S_{n} = x_{1}+ x_{2} + x_{3} + \cdots + x_{n}.$

Convergent SeriesA series is said to be convergent if the sequence $S_{1} , S_{2}, S_{3}, \ldots$ satisfies

$\lim_{n \rightarrow \infty} S_{n} = S,$

where $S$ is a finite value. So we can say that $\lim_{n \rightarrow \infty} \sum_{n=1}^{\infty} x_{n} = S$.

Divergent SeriesA series is said to be divergent if the sequence $S_{1} , S_{2}, S_{3}, \ldots$ does not converge, i.e. the limit of $S_{n},$ when $n$ approaches infinity, does not exist.

You can learn more about these in the wiki Infinite Sums.

Let us see some examples:

Show that the series $1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots$ is convergent.

The given series is actually an infinite geometric series with initial term $1$ and common ratio of $\frac{1}{2}$. But we know that an infinite geometric series converges if and only if $|r| < 1,$ where $r$ is the common ratio. So, the series is convergent. Let me prove that to make it clear to you:

$\begin{aligned} S &= 1+ \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \cdots \\ 2S &= 2 + 1 + \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \cdots \\ 2S - S &= 2 \\ S &= 2. \end{aligned}$

So our infinite series approaches a finite value of 2 and thus converges. $_\square$

Prove that the series $1+\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots$ is divergent.

The given series is actually an infinite harmonic series. Let $S$ denote the sum, then

$\begin{aligned} S &= 1 + \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} + \dfrac{1}{5} + \dfrac{1}{6} + \dfrac{1}{7}+ \dfrac{1}{8} + \cdots \\ &> 1+ \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{8} + \dfrac{1}{8} + \dfrac{1}{8} + \cdots \\ &= 1+ \dfrac{1}{2} +\dfrac{1}{2} + \dfrac{1}{2} + \cdots, \end{aligned}$

which approaches infinity. So, the series diverges. $_\square$

## Conditionally Convergent Series

Consider a series $a_{1} + a_{2} + a_{3} + a_{4} + \cdots$.

The above series is said to converge conditionally if and only if the series $\sum_{n=1}^{\infty} a_{n}$ converges and the series $\sum_{n=1}^{\infty} |a_{n}|$ diverges.

Let us see some examples regarding conditionally convergent series:

Evaluate the series

$\sum_{n=1}^{\infty} \left(\dfrac{1}{n} - \dfrac{1}{n-\frac{1}{2}}\right).$

We have

$\begin{aligned} \sum_{n=1}^{\infty} \left(\dfrac{1}{n} - \dfrac{1}{n-\frac{1}{2}}\right) &=\sum_{n=1}^{\infty} \left(\dfrac{2}{2n} - \dfrac{2}{2n-1}\right) \\ &= -2 \sum_{n=1}^{\infty} \left(\dfrac{1}{2n-1} - \dfrac{1}{2n} \right) \\ &= -2 \left(1 - \dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{4} + \cdots \right) \\ &= -2 \ln(2) \\ &= - \ln(4). \end{aligned}$

Is this the correct answer? Obviously, yes. We haven't done any rearrangements, so according to Riemann series theorem, the answer is correct.

There is a general misconception and people tend to do like this:

$\begin{aligned} \sum_{n=1}^{\infty} \left(\dfrac{1}{n} - \dfrac{1}{n-\frac{1}{2}}\right) &= \sum_{n=1}^{\infty} \left(\dfrac{1}{n} - \dfrac{2}{2n-1}\right) \\ &= \left( 1+ \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} + \cdots \right) - 2 \left(1+ \dfrac{1}{3} + \dfrac{1}{5} + \dfrac{1}{7} + \cdots \right) \\ &= - \left(1 - \dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{4} + \cdots \right) \\ &= - \ln(2). \end{aligned}$

We are getting a different value. But why? Have we made any mistake? Yes, we have made a big mistake. According to Riemann series theorem, when the terms in a series are rearranged, the series converges to a different value (not necessarily) or it may even diverge. This is the reason why we got a different value.

So, keep in mind that when dealing with infinite series, you should never rearrange the terms in the series. $_\square$

Evaluate the value of the series $1-\frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots$.

The above series can be written as

$1-\dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{4} + \cdots = -\sum_{n=1}^{\infty} \dfrac{(-1)^{n}}{n}.$

But the Taylor expansion of $\ln(1-x)$ is $- \sum_{n=1}^{\infty} \frac{x^{n}}{n}$.

So, the above sum becomes

$1-\dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{4} + \cdots = -\sum_{n=1}^{\infty} \dfrac{(-1)^{n}}{n} = \ln\big(1-(-1)\big)= \ln(2).$

Is this the correct answer? Obviously, yes. We haven't done any rearrangements, so according to Riemann series theorem, the answer is correct. $_\square$

**Cite as:**Riemann Series Theorem.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/riemann-series-theorem/