Question

Determine whether the sequence converges or diverges. If it converges, give the limit.
48, 12, 3, 3/4, ...

Answer 1

Converges; 64

Diverges

Converges; -4080

Converges; 0

I say converges at -4080

Answer 2

First we notice that this is a geometric series since the sequence is in the form:\[\bf ar^{n-1} \rightarrow 48\left( \frac{ 1 }{ 4 } \right)^{n-1}\]The geometric sequence converges if \(\bf |r| < 1\), i.e the common ratio is between -1 and 1, and diverges otherwise. Here we notice that the common ratio is:\[\bf 48\left( \frac{ 1 }{ 4 } \right)^{n-1} \implies r=\frac{ 1 }{ 4 } < \ 1\]Hence the series converges. Let's now find the limit of the sequence:\[\bf \lim_{n \rightarrow \infty}a_{n}=\lim_{n \rightarrow \infty}48\left( \frac{ 1 }{ 4 } \right)^{n-1}=48\lim_{n \rightarrow \infty}\frac{ 1 }{ 4^{n-1} }=0\]

Answer 3

@kjuchiha

Answer 4

Wow thank you! I can use the sequence formula u gave me for upcoming problems too! Tysm