Community post
Math Contest #26 Alternative Solution
As requested by @lallo I present you how I would have solved this.
This solution is a bit more complex and longer then the one presented in the Results Post, however it still kind of uses the same technique.
As a reminder the problem was to find the limit of the following converging sum:
And here my solution:
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Interesting that also you end up with an other series to solve, and exactly the same series I got too. Probably with (n^3)/(2^n) one get two "sub" series to solve and so on.
It seems like any such polynomial divided by a power can be reduced recursively to only get geometric series. This should also work for 3^n, 4^n e^n, … an for n³, n⁴, …
It could be interesting to see what happens with rational exponents like n^½.
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