Trivial Summation Challenge

Question

Write the repeating decimal $1.367367367367...$ in the form $a/b$ where both $a$ an $b$ are positive whole numbers, but do this using a converging infinite sum.

Answer 1

Rewrite your numer as $$1+0.367367\dots$$ and notice that the latter term can be rewritten as $$\sum_{n\geq 1}\frac{367}{10^{3n}}.$$ Now notice that this is nothing but an infinite geometric series. Conclude.

Answer 2

And, ignoring the requirement for an infinite sum, if $x=1.367367367367...$, then $1000x = 1367.367367367... $ so $999x = 1366$ so $x = \dfrac{1366}{999} $.