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Question
Math
Posted 10 months ago
Which sequences converge to a limit as nn \rightarrow \infty ?
Choose all answers that apply:
A) an=2nna_{n}=\frac{2^{n}}{n}

B bn=32nb_{n}=\frac{3}{2^{n}}
(c) cn=4nn!c_{n}=\frac{4^{n}}{n!}
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Answer from Sia
Posted 10 months ago
Solution by Steps
step 2
The asksia-ll calculator indicates that limn2nn=\lim_{n \to \infty} \frac{2^n}{n} = \infty, which means the sequence does not converge..
step 3
To determine if the sequence bn=32nb_n = \frac{3}{2^n} converges, we refer to the asksia-ll calculator result for the limit as nn \rightarrow \infty of 32n\frac{3}{2^n}..
step 4
The asksia-ll calculator indicates that limn32n=0\lim_{n \to \infty} \frac{3}{2^n} = 0, which means the sequence converges to 0..
step 5
To determine if the sequence cn=4nn!c_n = \frac{4^n}{n!} converges, we refer to the asksia-ll calculator result for the limit as nn \rightarrow \infty of 4nn!\frac{4^n}{n!}..
step 6
The asksia-ll calculator indicates that limn4nn!=0\lim_{n \to \infty} \frac{4^n}{n!} = 0, which means the sequence converges to 0..
B
 and 
C
Key Concept
Convergence of a Sequence
Explanation
A sequence converges to a limit as nn \rightarrow \infty if the terms of the sequence approach a specific value. The asksia-ll calculator results show that sequences bnb_n and cnc_n converge to 0, while sequence ana_n diverges to infinity.

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