Are the fractional parts of powers of $\pi$ divergent?

Question

Let us define $a_n$ as the fractional part of $\pi^n$.

In other words, define $a_n=\pi^n-\lfloor \pi^n \rfloor$.

Then, does the following limit exist? $$\lim_{n \to \infty}a_n$$Intuitively, it appears that it does not exist, though it does have a lower bound of $0$ and an upper bound of $1$. However, I have no idea how to proceed with a proof. Any help would be appreciated.

Answer 1

The sequence $\{a_n\}$ does not converge.

Furthermore, this sequence is said to be equidistributed or uniformly distributed, which means that the proportion of terms falling in any subinterval of $[0,1]$ is proportional to the length of the subinterval.

This is actually a specific case, of the more general question:

If $a_n=x^n-\lfloor x^n \rfloor$, for what values of $x$ is the sequence $\{a_n\}$ equidistributed?

Hardy showed that the this sequence is equisitributed for almost all values of $x$. The set of numbers such that it is not equidstributed, has Lebesgue measure of zero, and are called Pisot numbers. It includes $\phi =\frac{1+\sqrt{5}}{2}$ and $1+\sqrt{2}$.

Further details, visualizations and links can be found at the Wolfram Mathworld site.

Note

For those who may be new to studying sequences that involve fractional parts, be aware that curly braces $\{ \cdot \}$ are (unfortunately) used to convey two different concepts:

• to define a sequence $\{a_n\} = \{a_0,a_1,a_2,a_3,\ldots\}$;
• to define a function, where $\{x\} = x - \lfloor x \rfloor $, the fractional part of $x$.

Which usage is implied, can nearly always be deduced from context.