# Prove that for all x > 0, ∑ (-1)^n * x^(2n+1) / [(2n+1)*n!] < 0?

Prove that for all x > 0

∑ (-1)^n * x^(2n+1) / [(2n+1)*n!] < 0

n=1

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• Anonymous

I think I got it.

Let f(x) = your infinite series. You can show that f(x) converges for all x>0 using the ratio test.

I'm pretty sure f(x) converges absolutely for all x>0 (you might have to verify this on your own). If it does, then take the derivative term by term to get

f'(x) = ∑ (-1)^n * x^(2n) /n!

= ∑ (-1)^n * (x^2)^n /n!

= ∑ (- x^2)^n /n!

Remember: you sum runs from n = 1 to infinity. Change it to n = 0 and subtract 1 from it to get

f'(x) = ( ∑ (- x^2)^n /n! ) - 1.

You can do this because the expression (- x^2)^n /n! is equal to 1 when n =0 (from changing the lower limit of your sum), and so adding 1 to the sum is balanced out by the -1 at the end.

Now the sum is just the infinite series for e^(- x^2), so that your derivative is

f'(x) = e^(- x^2) - 1.

Since f'(x) < 0 for all x > 0, your original function is decreasing for all x > 0.

Since f(0) = 0 (original function), and since f(x) is decreasing for x > 0, you can safely conclude (I think) that f(x) < 0 for all x > 0.

I haven't done problems lke this is over a decade, so you might want to fill in the gaps/verify some claims I made.

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• Anonymous