A Geometric Series...?

Note that $\frac{60}{60^n-1} = \sum_{k=1}^{\infty} \frac{1}{60^{nk-1}}$

Writing out the base- $60$ expansions for the first few values of $n$ , this is

$\begin{aligned} \frac{60}{60-1} &= 1.11111\ldots_{60} \\ \frac{60}{60^2-1} &= 0.10101\ldots_{60} \\ \frac{60}{60^3-1} &= 0.01001\ldots_{60} \end{aligned}$

We are interested in the $60^{th}$ digit of the expansion, which is the $59^{th}$ digit after the sexagesimal point. Each $\frac{60}{60^n-1}$ term contributes a $1$ in this position if we can write $59$ in the form $nk-1$ for some positive integer $k$ - that is, if we can write $60=nk$ .

There are exactly as many of these $n$ as there are divisors of $60$ , that is, $12$ , and because we are only adding $60$ terms, there is no risk of interference from other digits; hence the answer is $\boxed{12}$ .

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These expansions work in any base; for example, the decimal expansions of $\frac{10}{9}$ , $\frac{10}{99}$ , and $\frac{10}{999}$ match (in terms of digits) the expansions listed above.

Write the given expression in the expanded form as,

$\\\displaystyle\sum_{n=1}^{60}\dfrac{60}{60^n-1}=\sum_{n=1}^{60}\dfrac{60^{1-n}}{1-60^{-n}}=\sum_{n=1}^{60}\sum_{k=1}^\infty\bigg(60^{1-n}\cdot60^{-n(k-1)}\bigg)=\sum_{n=1}^{60}\sum_{k=1}^\infty60^{1-nk}=\sum_{k=1}^{\infty}\sum_{n=1}^{60}60^{1-nk}\\$

We are interested in the value of the $60^{\text{th}}$ digit, which is here nothing but the coefficient of $60^{-59}$ in the expansion. If we can write $nk=60$ for some positive integers $n$ & $k$ , there are exactly as many of these $(n,k)$ as there are divisors of $60=2^2\cdot3\cdot5$ , that is, $3\cdot2\cdot2=12$ . Hence, the value of the $60^{\text{th}}$ digit of the given expression, when written in base $60$ , is $\boxed{12}$ . $\blacksquare$