In this problem you will consider the convergent series 1/2 + 1/6 + 1/12 + 1/20

Question

In this problem you will consider the convergent series 1/2 + 1/6 + 1/12 + 1/20 + ...., where the general term is 1/n(n + 1). (a) Recall from lecture that writing an infinite sum really represents the limit of the sequence of partial sums, if that limit exists. Here the partial sum of the first n term is S_n: = 1/2 + 1/6 + 1/12 + 1/20 + ... + 1/n(n + 1) Calculate the first several values of S_n, up to S_4 at the least (b) From your data in part (a), do you notice any patterns? Try to guess a general formula for S_n. (c) Notice (and verify) that 1/n(n +1) = 1/n - 1/n + 1. Plug this in for each term of S_n. If you keep track carefully of cancelation, you should now be able to prove your formula. (d) Finally, prove that lim S_n = 1. This is therefore the value of the infinite sum.

Explanation / Answer

General Term of the series is 1/n(n+1)

(a) Sn = 1/2 + 1/6+ 1/12 + 1/20 + ....+ 1/n(n+1)

s1 =1/2

s2 = 2/3

s3 = 2/3 + 1/12 = 3/4

s4 = 3/4 + 1/20 = 4/5

s5 = 4/5 + 1/30 = 5/6

(B) yes here we are noticing a pattern. We can see that sn = n/(n+1) is there when we sum al these numbers

(C) Here

Tn = 1/n(n+1) = 1/n - 1/(n+1)

so, S1 = 1/1.2 = 1/1 - 1/2

S2 = (1/1 - 1/2) + ( 1/2 - 1/3)

S3 = (1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4)

S4 = (1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + (1/4 - 1/5)

Sn = ( 1 -1/2) + (1/2 - 1/3) + (1/3 - 1/4) + (1/4 - 1/5) + .....................{1/n - 1/(n+1)}

so al the terms will cancel out and

Sn = 1 - 1/(n+1)

Sn = n/(n+1)

(d) FOr n =

Sn ( n -> ) = 1 - 1/ (n+1) = 1

S = 1

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