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 Erdos

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

  • Bounty is low for the amount of thinking and effort required. At least $50 probably makes sense

    • Im not really in a sich to throw money around but I raised it to twenty. I don't really need this to be answered I'm just kind of curious.

  • Erdos Erdos

    Do we assume that x is a number bigger than or equal 1?

    • It might work even if x is less than one, but I was assuming x was a natural number greater than 1.

    • It works for all real numbers greater than one as far as I can tell.

    • Correction: It seems to work for x ∈ (-∞, -1) U (1, ∞) not including +-1. In other words it seems to exist as long as the series diverges as k goes to infinity.


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Erdos Erdos
  • Erdos Erdos

    This took me a couple of hours to figure out. I was hesitant to respond, but since this appears to be the first question you've asked here, I decided to provide an answer. Please consider offering higher bounties to acknowledge the time spent by those answering your questions.

    • Sorry about the lower bounty. I'm a college student who's currently in debt, so money is kind of tight and this was more of curiosity than a need to know. Thanks for the explanation though. Hope you at least enjoyed solving it.

    • Erdos Erdos

      It's alright. I am glad I was able to help. You may enroll in the website's affiliate program and make money by introducing it to your friends/classmates at school. I personally make about a couple of hundred dollar each month. Here is the link: https://matchmaticians.com/affiliates

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