Determine if the series converges or diverges. If the series converges find its sum ∑n=1∞6(4n−1)(4n+3)

Question

Determine if the series converges or diverges. If the series converges find its sum  ∑n=1∞6(4n−1)(4n+3)

Sadie Eaton · Accepted Answer

Given:  Sn=∑n=1∞6(4n−1)(4n+3)  From the above expression, substitute n=1 to find the first term of the series.  6(4×1−1)(4×1+3)=63×7  =621  Substitute n=2 to find the 2nd term in the series.  6(4×2−1)(4×2+3)=67×11  =677  Substitute n=3 to find the 2nd term in the series.  6(4×3−1)(4×3+3)=611×15  =6165  The series can be written as,  621,677,6165,...,6(4n−1)(4n+3)  Let the value of n tends to infinity, the denominator also tends to infinity, and the last term tends to zero. Thus, the above series is converging.  Sn=∑n=1∞6(4n−1)(4n+2)  =∑n=1∞(64)(4n+3−(4n−1)(4n−1)(4n+3))  =(32)∑n=1∞(4n+3(4n−1)(4n+3)−(4n−1)(4n−1)(4n+3))  =(33)∑n=1∞(1(4n−1)−14n+3)  Substitute values of n for each term in the above expression.  Sn=32(14−1−14+3)+32(18−1−18+3)+32(112−1−112+3)+...+32(14n−1−14n+3)  =32[(13−17)+(17−111)+(111−115)+...+(14n−1−14n+3)]  =32[13−17+17−111+111−115+...−14n−1+14n−1−14n+3]  =32[13−14n+3]  =32[4n+3−3(4n+3)3]  =12[4n4n+3]  =2n4n+3  Thus, the sum of the series is 2n4n+3

Jeffrey Jordon · Accepted Answer

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Determine if the series converges or diverges. If the series converges find its sum sum_{n=1}^inftyfrac{6}{(4n-1)(4n+3)}

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