Question

Please read the question and follow its directions as the way I solve this problem is important.

Determine whether the series is convergent or divergent by expressing Sn as a telescoping sum. Me 2 02 + 4n + 3 n = 1 convergent divergent If it is convergent, find its sum. (If an answer does not exist, enter DNE.)

Please show ALL work and also give an explanation to why it is convergent or divergent, that is also important.

Answer 1

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Thanks

Solution n²+4nt3 n-1 2 We have I dans n²+4n+3 (n+1)(n+3) n=1 n1 using bastial fraction decomposition we will wife if into telescoping B 2 + (n+1) +3) (+1) (n+3 2 = (1+3) A + B (n+1) 2 = n(A+B) † (3 A+B) 3A + B = 2 X + B co A =-B -2B=2 B=-| A =

we can write if an 2 62 +40 +3 6+0 M +1 (n+3) Therefore the partial sum tan Sn = at a ta ta t.. = ( = x + ( 73 - * + + (*) t ca + 1 4 (ht + th +2) +(() n+2 h+3 Sn 은 1 + 1 n+3 I nt2 ویر) 3. e 5 I Sn that 6 (n+2) (n+3)

Therefore Sn = lim E 2 n²+ 40+ no D1 = lim (-btz -63) neo value is finite 6 so series is conver onvergent 5 6 Sum of series is