r/askmath 9d ago

Discrete Math Simplify sum

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Friend gave this "easy" sum for Olympiad practice, can't solve it even though I already tried relating it to e's Taylor series done harmonic stuff etc, any help?

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u/uTRexAap 9d ago

havent taken any calc so im terrible at this but i got it down to this

u/ZellHall 8d ago

Great job!

u/uTRexAap 8d ago

Thanks but why?

u/BubbhaJebus 6d ago

Then reindex it.

u/uTRexAap 6d ago

What does that mean?

u/BubbhaJebus 6d ago

Rewrite it so the index of summation ranges from 1 to n-1 instead of 2 to n. It's a little simpler that way.

u/uTRexAap 6d ago

You can do that???????

u/BubbhaJebus 6d ago

Yes: SUM(i = 2 to n) f(n) = SUM(i = 1 to n-1) f(n+1)

u/uTRexAap 6d ago

At that point why not 0 to n-2 ?

u/Bounded_sequencE 9d ago

What exactly is the assignment? Also, please provide a link to the official olympiad paper, to ensure this is is a legit question.

u/Equivalent_Dingo_683 9d ago

friend made it up and claims to have a closed solution but isn't giving it

u/Bounded_sequencE 9d ago

Chances are high you have been trolled successfully^^

You also did not answer the question what exactly the assignment is.

u/AppropriateCar2261 9d ago

You probably have a typo. I gave this sum to Wolfram Mathematica, and even it couldn't find a simpler expression.

u/Bounded_sequencE 9d ago edited 9d ago

It's fairly easy to find an upper estimate to show the series tends to zero as "n -> oo". That's about the only "nice" pattern I've been able to find so far.

u/Equivalent_Dingo_683 9d ago

nope just confirmed it's not a typo

u/uTRexAap 8d ago

what is the intended sum?

u/veryjewygranola 9d ago

If you define your sum up to n to be S(n), and look at the ratios of S(n)/S(n-1), you see it approaches 1/2 rather quickly, so we can approximate the asymptotic behavior of S(n) as 2^(-n+1) (we do n+1 since S(2) = 1/2).

A closed form or exact expression for this sum is unlikely.

u/Mission_Rice3045 9d ago

Hmm, what happens if you take the log of the sum terms? This would turn the factorial into a sum

u/Bounded_sequencE 9d ago

Logarithms are not linear -- applying "ln(..)" to the entire sum does not carry over to the argument.

u/Mission_Rice3045 9d ago

Yes, but (...) = e^{ln(...)}. Like I said this allow you to write the factorial as a sum. Don't know if this will meaningfully simplify this summation but it is something commonly done.

u/Equivalent_Dingo_683 8d ago

Idt it's going anything :(

u/BubbhaJebus 6d ago

SUM(L = 1 to n-1)(L+1)!^(L-n)