Divergent Series Yield Valuable Results

Mathematicians have traditionally dealt with convergent series and shunned divergent ones. But, long ago, astronomers found that divergent expansions yield valuable results. If these so-called asymptotic expansions are truncated, the error is bounded by the first term omitted. Thus, by stopping just before the smallest term, excellent approximations may be obtained.

Astronomical Series

Many of the series that astronomers have used do not converge. Henri Poincaré pointed out the difference in viewpoint between geometers and astronomers — or between pure and applied mathematicians. He considered first the series

\displaystyle \sum_{n=0}^\infty \left(\frac{1000^n}{n!}\right)x^n

This is the Taylor series for {\exp(1000 x)}, and it is convergent for all values of {x}. However, it is a nightmare to compute, as the terms grow rapidly before decreasing to zero.

In contrast, the series

\displaystyle \sum_{n=0}^\infty \left(\frac{n!}{1000^n}\right) \frac{1}{x^n}

diverges for all nonzero {x}. However, the terms decrease rapidly at first, and only after {n} exceeds {1000x} do they begin to increase. The first twenty terms of the two series are shown for {x=0.01}.


The first twenty terms of the sequences {10^n/n!} (left) and {n!/10^n} (right).

Convergence & Divergence

For the infinite series

\displaystyle S := a_1 + a_2 + a_3 + a_4 + \dots

we define the sequence of partial sums

\displaystyle S_1 = a_1 \,, \qquad S_2 = a_1 + a_2 \,, \qquad S_3 = a_1 + a_2 + a_3 \,, \quad \dots \quad \,.

The series {S} is convergent if the sequence of partial sums has a finite limit. If the sequence {\{S_n\}} has no such finite limit, the series {S} is divergent. The sequence may oscillate regularly or irregularly or tend to an infinite limit.

Obviously, we cannot compute the sum of a divergent series in the usual way. However, there are several “summation methods” for associating a number with a divergent series. Leonhard Euler was a maestro of divergent series, and he obtained many extraordinary results using ingenious methods.

An Example

One of Euler’s examples was the divergent series

\displaystyle S := 1 - 1! + 2! - 3! + 4! - 5! + \dots

or, more explicitly

\displaystyle S := 1 - 1 + 2 - 6 + 24 - 120 + \dots \,.

It is clear that the partial sums oscillate between positive and negative values with ever-increasing amplitude, so the series is divergent. Nevertheless, Euler assigned the value {S \approx 0.596347} to the sum of the series. He arrived at this value using no fewer than five different summation methods. We will look at one of these, based on the solution of a differential equation. Let us consider the linear first-order ordinary differential equation

\displaystyle x^2 y^\prime + y = x \,, \qquad y(0) = 0 \,.

Solution in Series

There are several ways to solve this differential equation. First, let’s seek a solution in the form of a power series

\displaystyle y(x) = \sum_{n=0}^\infty a_n x^n \,.

Substituting this into the equation, and equating the coefficients of different powers of {x}, we find

\displaystyle a_0 = 0\,, \qquad a_1 = 1\,, \quad \dots \quad a_{n+1} = - n a_n \,, \quad \dots

So the series expansion becomes

\displaystyle S(x) = x - 1! x^2 + 2! x^3 - 3! x^4 + \dots \,.

Setting {x = 1}, this is just Euler’s series that we started with above:

\displaystyle S(1) = 1 - 1! + 2! - 3! + \dots \,.

Applying the ratio test to the series {S(x)} we have

\displaystyle \left| \frac{a_{n+1}}{a_n}\right| = n | x |

which has the limit {+\infty} for all nonzero {x}, so the series diverges unless {x=0}. In particular, the value {S(1)} is not defined by the series.

An Integrating Factor

Another way to solve the differential equation is by means of an integrating factor. We write the equation as

\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x} + \left(\frac{1}{x^2}\right) y = \frac{1}{x} \,.

Now multiply by the exponential of the integral of the coefficient of {y}:

\displaystyle P(x) = \exp\left( \int \frac{1}{x^2}\, \mathrm{d} x \right) = \exp\left( - \frac{1}{x} \right) \,.

The equation becomes

\displaystyle \frac{\mathrm{d}}{\mathrm{d} x} [ e^{-1/x} y ] = \frac{1}{x} e^{-1/x} \,,

and the solution follows:

\displaystyle y(x) = e^{1/x} \int_0^x \frac{1}{u} e^{-1/u} \,\mathrm{d} u

A non-obvious change of variable {v = (x-u)/xu} or {u = x / (1 + xv)} converts this to

\displaystyle y(x) = \int_0^\infty \frac{x}{1+xv} e^{-v} \,\mathrm{d} v

This can be expressed in terms of the standard exponential integral Ei(x), but we will not worry about that. Expanding the component {1/(1+xv)} by the binomial theorem and integrating term by term, we get

\displaystyle y(x) = x - 1! x^2 + 2! x^3 - 3! x^4 + \dots \,,

the same series expansion that we obtained above. The series

\displaystyle S(x) = x - 1! x^2 + 2! x^3 - 3! x^4 + \dots \,.

is an asymptotic expansion. It is divergent but, if we stop just before the smallest term, we get an approximation to the true value.

Numerical Results

Suppose {x=0.1}; then the terms decrease until {n=10} and then increase. Summing the first ten terms and comparing it to the integral solution, we get

\displaystyle S(0.1) \approx 0.0915456 \qquad\qquad y(0.1) = 0.0915633 \,,

accurate to three significant figures. Now, taking {x=0.01}, the terms decrease until {n=100}. Summing the first 99 terms and comparing it to the integral solution, we get

\displaystyle S(0.1) \approx 0.00990194 \qquad\qquad y(0.1) = 0.00990194 \,,

which is accurate to (at least) six significant digits.

Let’s recap: we have a solution of the ode in terms of a divergent series

\displaystyle S(x) = x - 1! x^2 + 2! x^3 - 3! x^4 + \dots \,,

and a solution in terms of an integral

\displaystyle y(x) = \int_0^\infty \frac{x}{1+xv} e^{-v} \,\mathrm{d} v \,.

The integral is well defined and, for {x=1}, it has the value {0.596347362323194 \dots}. This is the value that Euler associates with the divergent series

\displaystyle S = 1 - 1! + 2! - 3! + \dots \sim 0.596347362323194 \,.


{\bullet} Étienne Ghys, 2017: A Singular Mathematical Promenade. ENS Éditions, 306pp. ISBN: 978-2-8478-8940-6.

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