Determine if the following series converge. If the series converges, calculate its value. Justify your answer. ∑n=1∞n2n!(log⁡(2))n

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Determine if the following series converge. If the series converges, calculate its value. Justify your answer.   ∑n=1∞n2n!(log⁡(2))n

krolaniaN · Accepted Answer

Given:  ∑n=1∞n2n!(log⁡(2))n  Apply the series ratio test to find the series is convergence or divergence.  If the value of L comes greater  than 1 is called divergence. and the value of L comes less than 1 is called convergence.  L=limn→∞|an+1an|  =limn→∞|(n+1)2(n+1)!(log⁡2)n+1n2n!(log⁡2)n|  =limn→∞|(n+1)n2(log⁡2)|  the value of L comes less than 1 is called convergence.  Find the value of given series.  ∑n=1∞n2n!(log⁡2)n=121!(log⁡2)1+222!(log⁡2)2+323!(log⁡2)3+424!(log⁡2)4+...  =1(log⁡2)+2(log⁡2)2+32(log⁡2)3+43(log⁡2)4+...  =n2(log⁡2)n!(1−log⁡2)  which is the solution.

Jeffrey Jordon · Accepted Answer

Answer is given below (on video)

Determine if the following series converge. If the series converges, calculate its value. Justify your answer. sum_{n=1}^inftyfrac{n^2}{n!(log(2))^n}

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