Question

(1 point) For each of the following, carefully determine whether the series converges or not.\\ (a) $\sum_{n=1}^{\infty} \frac{n^3}{n!}$ \\ A. converges\\B. diverges\\(b) $\sum_{n=1}^{\infty} \frac{(-1)^{n-1} n^3}{4^n}$ \\ A. converges\\B. diverges\\(c) $\sum_{n=2}^{\infty} \frac{4}{\ln n^3}$ \\ A. converges\\B. diverges\\(d) $\sum_{n=1}^{\infty} \frac{n(n+2)}{\sqrt{n^3 + 4n^2}}$ \\ A. converges\\B. diverges

          (1 point) For each of the following, carefully determine whether the series converges or not.\\
(a) $\sum_{n=1}^{\infty} \frac{n^3}{n!}$ \\
A. converges\\B. diverges\\(b) $\sum_{n=1}^{\infty} \frac{(-1)^{n-1} n^3}{4^n}$ \\
A. converges\\B. diverges\\(c) $\sum_{n=2}^{\infty} \frac{4}{\ln n^3}$ \\
A. converges\\B. diverges\\(d) $\sum_{n=1}^{\infty} \frac{n(n+2)}{\sqrt{n^3 + 4n^2}}$ \\
A. converges\\B. diverges

(1 point) For each of the following, carefully determine whether the series converges or not.

(a) ∑n=1^∞(n^3)/(n!)

A. converges
B. diverges
(b) ∑n=1^∞((-1)^n-1 n^3)/(4^n)

A. converges
B. diverges
(c) ∑n=2^∞(4)/(ln n^3)

A. converges
B. diverges
(d) ∑n=1^∞(n(n+2))/(√(n^3 + 4n^2))

A. converges
B. diverges

Added by Salvador S.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Adi S

09/14/2023

Step 1

The ratio test states that if the limit of the absolute value of the ratio of consecutive terms is less than 1, then the series converges. Show more…

Show all steps

Thanks for your feedback!

Can you please explain this question in detail. converges or not. 8 n3 (a) n! n=1 . A. converges B. diverges 8 (-1)-1n3 (b) 2 4n n=1 . A. converges B. diverges 8 4 (c) Z ln n3 n=2 A. converges . B. diverges 8 n(n+2) (d) 2 Vn3+4n2 n=1 A. converges B. diverges