Sum of Convergent SeriesDate: 09/25/1999 at 23:49:18 From: Abby Subject: Sum of a convergent series Could someone tell me how to find the sum of a convergent series such as these: infinity Sum 1/[(k+1)(k+3)] k = 0 infinity Sum [(25/10^k) - (6/100^k)] k = 0 Thanks in advance. Date: 09/26/1999 at 08:14:24 From: Doctor Anthony Subject: Re: Sum of a convergent series > infinity > Sum 1/[(k+1)(k+3)] > k = 0 1 A B ---------- = --- + --- (k+1)(k+3) k+1 k+3 1 = A(k+3) + B(k+1) When k = -1: 1 = 2A and so A = 1/2 k = -3: 1 = -2B and so B = -1/2 1 1 1 Therefore ---------- = ------ - ------ (k+1)(k+3) 2(k+1) 2(k+3) Our series is therefore the sum of the following terms: 1/[2(1)] - 1/[2(3)] + 1/[2(2)] - 1/[2(4)] + 1/[2(3)] - 1/[2(5)] + 1/[2(4)] - 1/[2(6)] + : : and so on to infinity. We can see that terms will start to cancel between lines except the first terms on the first and second lines. So the sum to infinity is: 1/[2(1)] + 1/[2(2)] = 1/2 + 1/4 = 3/4 > infinity > Sum [(25/10^k) - (6/100^k)] > k = 0 We could split this into two series 25[1 + 1/10 + 1/10^2 + ...] - 6[1 + 1/100 + 1/100^2 + ...] These are both geometric series, and using the formula a/(1-r) as sum to infinity 25/[1-(1/10)] - 6/[1-(1/100)] = 250/9 - 600/99 = 2150/99 = 21.717171... - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
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