# Wierdly Rational Fractions

Prove that as $k$ gets bigger ${\frac{\sum_{n=1}^{k}{nx^{n-1}}}{\sum_{n=1}^{k}{nx^{k-n}}}}$ approaches $k(1-\frac{1}{x}) -\frac{1}{x}$ when $x∈(\infty ,-1)∪(1,\infty )$ , and find an equation for the error between the two using $k$ and $x$ as variables. I understand that $\lim_{k\rightarrow \infty}f(k)$ just evaluates to $\infty$, which is why I tried to word the question a little differently. I was putting random fractions into my calculator, because I'm wierd, and I noticed that $\frac{987654321}{123456789}$ is abnormaly close to 8. At first I assumed this was just a coincidence, but after playing around with different amounts of digits and different number bases, I noticed that very similar patterns emerged. I also noticed that as the number of digits used grew larger, it approximated the patterns much more closely. I came up with the formula above where $x$ is the number base you are using and $k$ is the highest "digit" in the fraction. The quotation marks around "digit" are because anything past the ($x$ - 1)th digit in any given number system of base $x$ is actually two or more digits, so technically when $k=10$ and $x=10$ the numerator will have 11 digits (10987654321) the highest of which is only 9. I assume there's a simple reason for this phenomenom, especially because as the denominator gets bigger, the added numbers start to matter less and less (multiplying by 10 would achieve a very similar effect), but I'm too stupid to figure it out, so all help is appreciated. This question has probably been asked before, but I can't find a good answer anywhere online.I'll accept a logical proof instead of a mathematical one if it makes sense.

• Erdos
+2

This is a computationally heavy problem and requires quite a bit of work. I think you should offer a higher bounty.