Type your question here Given the series Find the first six partial sums and wri
ID: 2847818 • Letter: T
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Given the series Find the first six partial sums and write them in a sequence, ,s2,s3, s4,s5,s6,...,sn,..., then describe the sequence of partial sums using as many adjectives as you can (increasing, decreasing, monotonic, bounded, alternating, etc.). Is the sequence of partial sums convergent or divergent? If the sequence is convergent, guess the limit of this sequence. Remember, the main mathematical idea we are trying to conceptualize here is that the sum of the series is the limit of the sequence of its partial sums. Take time to reflect on this, it is not an easy concept! Use partial fraction decomposition to find a formula for the nth partial sum of the series, that is sn = ai = a1 + a2 + a3 + .... +an-1 + an. This series should telescope down to four terms, the last term being the nth term. Use the expanded form to find the sum of the series, S, where S = lims n->infinity Sn.Explanation / Answer
(a)
S1 = 8/8 = 1
S2 = 1+8/15 = 23/15
S3 = 23/15 + 8/24 = 28/15
S4 = 28/15+8/35 = 220/105 = 44/21
S5 = 44/21 +8/48 = 95/42
S6 = 95/42+8/63 = 301/126
it is incresing, monotonic,bounded above,
(b)
8/(n^2+4n+3) = 8/(n+1)(n+3) = 4[1/(n+1) - (n+3)]
T1 = 4[1/2 - 1/4]
T2 = 4[1/3 - 1/5]
T3 = 4[1/4 - 1/6]
.
.
.
Tn = 4[1/(n+1) - 1/(n+3)]
adding all these terms gives
Sn = 4[1/2 +1/3 - 1/(n+2) - 1/(n+3)
= 4[5/6 -(2n+5)/(n+2)(n+3)]
=> Sn is convergent
(c)
Sn = 4[1/2 +1/3 - 1/(n+2) - 1/(n+3)
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