Σn=(-n)n/(2n+1)!

Question

Σn=(-n)ⁿ/(2n+1)!

What's the exact sum and how many terms necessary to estimate the sum to three decimal places? I know it's convergent

Answer 1

This series converges because lim n → ∞ of (nⁿ/(2n + 1)!) = lim n → ∞ of (n·n·n·...n/1·2·3·...n.(n+1)(n+ 2)...2n(2n+1)· = 0.

Now a_{n + 1} = (n + 1)^{n + 1}/(2n + 3)! < nⁿ/(2n + 1)! = a_n, becouse after we reduce we get (1+ 1/n)ⁿ < 2(2n +3) or

(1 + 1/n)ⁿ < 4n + 6 and this is true for any n = 1, 2, 3, ... because we know that 2 < (1 + 1/n)ⁿ < 3

(look please prove of lim n → ∞ of (1 + 1/n)ⁿ = e = 2.718....

So by Alternative series test our series converges

Answer 2

Hi Lily S.,

I don't know whether you are being asked to deliver an analytical result, or simply a numerical one. If the latter, don't waste time, just throw into your favorite spreadsheet program and tabulate. Four terms gets you -0.138; exact is around . I doubt that it's a rational number.

Now, the analytical result (i.e. by means of a higher math function) -- I'm sure it exists, but I'm not the mathematician to show you how.

-- Cheers, -- Mr. d.