Question #273831

find the general term of the sequence, starting with n = 1. Determine whether the sequence converges and if so find its limit. If the sequence diverges, indicate that using the checkbox.


3, 3/19, 3/19^2, 3/19^3...



1
Expert's answer
2021-12-01T17:22:41-0500

3, 3/19, 3/19^2, 3/19^3...

Given sequence is a GP whose first term (a)(a) is 3 and common ratio (r)(r) is 1/19.

nthn^{th} term of a GP is given by: an=a.rn1a_n=a.r^{n-1}

an=319n1a_n=\frac{3}{19^{n-1}}

limnan=limn319n1=0\lim_{n\to\infty} a_n= \lim_{n\to\infty} \frac{3}{19^{n-1}}=0


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