# 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 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 terms could affect the sum.

## Definitions

Consider a sequence \(x_{1}, x_{2}, x_{3}, x_{4} \ldots\). Let \(S_{n} = x_{1}+ x_{2} + x_{3} + \ldots + x_{n}\).

Convergent Series :A series is said to be convergent if the sequence \(S_{1} , S_{2}, S_{3} \ldots \) is convergent i.e. \[\lim_{n \rightarrow \infty} S_{n} = S\] where \(S\) is a finite value.So we can say that \(\lim_{n \rightarrow \infty} \sum_{n=1}^{\infty} a_{n} = S\).

Divergent Series :A 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 at the wiki Infinite Sums.

Let us see examples regarding these.

## The series \(1 + \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \ldots \) is convergent.

We shall prove that the given series is convergent.

The given series is actually a Infinite Geometric Series with initial term \(1\) and with common ratio of \(\dfrac{1}{2}\). But we know that Infinite Geometric series converges iff \(|r| < 1\), where \(r\) is the common ratio. So, the series is convergent. Let me prove that to make it clear to you.

\[\begin{align*} S &= 1+ \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \ldots \\ 2S &= 2 + 1 + \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \ldots \\ 2S - S &= 2 \\ S &= 2 \end{align*} \]

So our infinite series approaches a finite value i.e. \(2\). So the series converges.

## The series \(1+\dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} + \ldots \) is divergent.

We shall prove that the given series is divergent.

The given series is actually a Infinite Harmonic Series. Let us denote the sum with letter \(S\).

\[\begin{align*} S &= 1 + \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} + \dfrac{1}{5} + \dfrac{1}{6} + \dfrac{1}{7}+ \dfrac{1}{8} + \ldots \\ &> 1+ \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{8} + \dfrac{1}{8} + \dfrac{1}{8} + \ldots \\ &= 1+ \dfrac{1}{2} +\dfrac{1}{2} + \dfrac{1}{2} + \ldots \end{align*}\]

We see that the sum \(1+ \dfrac{1}{2} +\dfrac{1}{2} + \dfrac{1}{2} + \ldots\) approaches infinity as it eventually increasing with a factor of \(\dfrac{1}{2}\). So, the series diverges.

## Conditionally Convergent series

Consider a series \(a_{1} + a_{2} + a_{3} + a_{4} + \ldots\).

The above mentioned 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.

## Example 1 : Evaluate the value of series

\[\sum_{n=1}^{\infty} \dfrac{1}{n} - \dfrac{1}{n-\dfrac{1}{2}}\]

\[\begin{align*} \sum_{n=1}^{\infty} \dfrac{1}{n} - \dfrac{1}{n-\dfrac{1}{2}} &=\sum_{n=1}^{\infty} \dfrac{2}{2n} - \dfrac{2}{2n-1} \\ &= -2 \left( \sum_{n=1}^{\infty} \dfrac{1}{2n-1} - \dfrac{1}{2n} \right) \\ &= -2 \left(1 - \dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{4} + \ldots \right) \\ &= -2 \ln(2) \\ &= - \ln(4) \end{align*}\]

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

There is a general mis-conception and people tend to do like this.

\[\begin{align*} \sum_{n=1}^{\infty} \dfrac{1}{n} - \dfrac{1}{n-\dfrac{1}{2}} &= \sum_{n=1}^{\infty} \dfrac{1}{n} - \dfrac{2}{2n-1} \\ &= \left( 1+ \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} + \ldots \right) - 2 \left(1+ \dfrac{1}{3} + \dfrac{1}{5} + \dfrac{1}{7} + \ldots \right) \\ &= - \left(1 - \dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{4} + \ldots \right) \\ &= - \ln(2) \end{align*} \]

We are getting a different value. But why? Have we done any mistake? Yes, we have done a big mistake. According to Riemann Series Theorem, when the terms in a series are rearranged the series converges to a different value (not necessary) 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.

## Example 2 : Evaluate the value of the series \(1-\dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{4} + \ldots \).

The above series can be written as

\[1-\dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{4} + \ldots = -\sum_{n=1}^{\infty} \dfrac{(-1)^{n}}{n}\]

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

So, the above sum becomes

\[1-\dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{4} + \ldots = -\sum_{n=1}^{\infty} \dfrac{(-1)^{n}}{n} = \ln(1-(-1))= \ln(2)\]

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

**Cite as:**Riemann Series Theorem.

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