Question

2. Determine if the series converges or diverges. Give reasons for your answer. If the series converges, find its sum. (a) $\sum_{n=1}^{\infty} \frac{n^n}{n!}$ (b) $\sum_{n=1}^{\infty} \frac{6}{(2n-1)(2n+1)}$ (c) $\sum_{k=1}^{\infty} (\sqrt{k}-1)^{2k}$ (d) $\sum_{k=1}^{\infty} \frac{2^k + 3^k}{4^k}$

Name: 2 determine if the series converges or diverges give reasons for your answer if the series converges find its sum a sumn1infty fracnnn b sumn1infty frac62n 12n1 c sumk1infty sqrtk 12k d s 81895
Uploaded: 2022-01-20T16:19:34-08:00
Duration: 4 min 14 s
Channel: Eduard Sanchez
Description: 2 determine if the series converges or diverges give reasons for your answer if the series converges find its sum a sumn1infty fracnnn b sumn1infty frac62n 12n1 c sumk1infty sqrtk 12k d s 81895

          2. Determine if the series converges or diverges. Give reasons for your answer. If the series converges, find its sum.
(a) $\sum_{n=1}^{\infty} \frac{n^n}{n!}$
(b) $\sum_{n=1}^{\infty} \frac{6}{(2n-1)(2n+1)}$
(c) $\sum_{k=1}^{\infty} (\sqrt{k}-1)^{2k}$
(d) $\sum_{k=1}^{\infty} \frac{2^k + 3^k}{4^k}$

$2 determine if the series converges or diverges give reasons for your answer if the series converges find its sum a sumn1infty fracnnn b sumn1infty frac62n 12n1 c sumk1infty sqrtk 12k d s 81895$

Added by Consuelo R.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Eduard Sanchez

01/20/2022

Step 1

Let $a_n = \frac{n^n}{n!}$. Then $$ \frac{a_{n+1}}{a_n} = \frac{(n+1)^{n+1}}{(n+1)!} \cdot \frac{n!}{n^n} = \frac{(n+1)^{n+1}}{(n+1)n^n} = \frac{(n+1)^n}{n^n} = \left(1 + \frac{1}{n}\right)^n $$ As $n \to \infty$, $\left(1 + \frac{1}{n}\right)^n \to e > 1$. Thus, Show more…

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2. Determine if the series converges or diverges. Give reasons for your answer. If the series converges, find its sum. (a) $\sum_{n=1}^{\infty} \frac{n^n}{n!}$ (b) $\sum_{n=1}^{\infty} \frac{6}{(2n-1)(2n+1)}$ (c) $\sum_{k=1}^{\infty} (\sqrt{k}-1)^{2k}$ (d) $\sum_{k=1}^{\infty} \frac{2^k + 3^k}{4^k}$