Determine whether the following series converge. If they do, find the sum. Justi

Question

Determine whether the following series converge. If they do, find the sum. Justify
your answers.

A) the sum from n=1 to infinity of [1 / (2n+1)(2n+2)]
B) the sum from n=1 to infinity of [3^n / (n-1)!]

please show all work and explain if you can so I can understand. Please write clearly.
Thank you!

Explanation / Answer

Start with the geometric series (convergent for |x| < 1). 1/(1 - x) = S(n=0 to 8) x^n Integrate both sides from 0 to x: -ln(1 - x) = S(n=0 to 8) x^(n+1)/(n+1) Re-indexing the sum up one unit: -ln(1 - x) = S(n=1 to 8) x^n/n. Let x = t^2: -ln(1 - t^2) = S(n=1 to 8) t^(2n)/n. --------------------------- By integration by parts, ? ln z dz = z ln z - z + C Therefore, ?(0 to x) -ln(1 - t^2) dt = ?(0 to x) - [ln(1 - t) + ln(1 + t)] dt = [(1 - t) ln(1 - t) - (1 - t)] - [(1 + t) ln(1 + t) - (1 + t)] {for t = 0 to x} = -(1 + t) ln(1 + t) + (1 - t) ln(1 - t) + 2t {for t = 0 to x} = -(1 + x) ln(1 + x) + (1 - x) ln(1 - x) + 2x ---------------------------- Integrate both sides of the previous series from 0 to x: -(1 + x) ln(1 + x) + (1 - x) ln(1 - x) + 2x = S(n=1 to 8) x^(2n+1)/[n(2n+1)] Finally, let x --> 1- -2 ln 2 + 0 + 2 = S(n=1 to 8) 1/[n(2n+1)] (L'Hopital's Rule is used in the middle term on the left side.) Therefore, S(n=1 to 8) 1/[n(2n+1)] = 2 - 2 ln 2. a[n] = n! / n^n r = lim(n->infinity) |a[n+1]/a[n]| = lim(n->infinity) (n+1)!/(n+1)^(n+1) * n^n/n! (n+1)!/(n+1)^(n+1) * n^n/n! = (n+1)!/n! * (1/(n+1)) * (n/(n+1))^n = (n/(n+1))^n r = lim(n->infinity) (n/(n+1))^n = 1/e Since 1/e < 1, the series converges.

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