Question

Test the series below for convergence using the Ratio Test:

∑[n=0 to [infinity]] ((-1)ⁿ 10²ⁿ⁺¹) / ((2n+1)!)

The limit of the ratio test simplifies to lim[n→[infinity]] Where f(n) = ___

Answer 1

Using the Ratio Test, the limit of the ratio of successive terms for the provided series is 0 as n approaches infinity, which means the series converges absolutely. The function f(n) for the Ratio Test is 10² / ((2n+3)(2n+2)).

Step-by-step explanation:

We are tasked to test the convergence of the series ∑[n=0 to ∞] ((-1)ⁿ 10²ⁿ¹) / ((2n+1)!) using the Ratio Test. The Ratio Test involves finding the limit of the absolute value of the ratio of successive terms of the series as n approaches infinity. For our series, this equates to:

lim[n→∞] |an+1/an| = lim[n→∞] |((-1)n+1 102(n+1)+1 / (2(n+1)+1)!) / ((-1)ⁿ 10²ⁿ¹ / (2n+1)!)|

This simplifies to:

lim[n→∞] |(-10² / (2n+3)(2n+2))| = lim[n→∞] (10² / ((2n+3)(2n+2)))

Since the limit of 10² / ((2n+3)(2n+2)) as n approaches infinity is 0, and 0 < 1, by the Ratio Test, the series converges absolutely. Therefore, the function f(n) in the Ratio Test is 10² / ((2n+3)(2n+2)).