Question

A sequence X is defined recursively as follows:
s1 = 100
sn+1 = 1/2 (sn + 1/3).
Find lim Sn (as n approaches infinity).


The formula I got for Sn = (50/ 3) * (n-1) * (1/2) ^(n-1). Also, I got the limit of Sn (as n approaches infinity) is 0. However, I'm not sure if this is correct or not.

Answer 1

\(\color{black}{\displaystyle s_1=100 }\)
\(\color{black}{\displaystyle s_{n+1}=\frac{1}{2}\left(s_n+\frac{1}{3}\right)}\)

like this?

Answer 2

yes

Answer 3

\(\color{black}{\displaystyle s_{n+1}=\left(\frac{3s_n+ 1}{6}\right).}\)

\(\color{black}{\displaystyle s_{2}=\left(\frac{301}{6}\right).}\)
\(\color{black}{\displaystyle s_{3}=\left(\frac{904}{36}\right).}\)
\(\color{black}{\displaystyle s_{3}=\left(\frac{904}{6^3}\right).}\)

etc, so you know intuitively.

Answer 4

oh, excuse me, that should be 2713, not 904.

Answer 5

We can do something like a Comparison Test, but as regards to convergence of a sequence (not series).

\(\color{black}{\displaystyle s_1=1000 }\)

\(\color{black}{\displaystyle s_{n+1}=\frac{2s_n}{3}}\)

\(\color{black}{\displaystyle s_{n+2}=\frac{\frac{2s_n}{3}}{3}=\frac{2s_n}{3^2}}\)

and then,
\(\color{black}{\displaystyle s_{n+k}=\frac{\frac{2s_n}{3}}{3}=\frac{2s_n}{3^k}}\)

Answer 6

basically, you keep dividing by 3, in the above case.
and thus, it will go to 0 with more and more terms.

Answer 7

So, for your case, where a smaller sequence is considered .... (you know)

Answer 8

We considered a larger sequence than yours, but we can also take a smaller one.

\(\color{black}{\displaystyle s_1=10 }\)

\(\color{black}{\displaystyle s_{n+1}=\frac{1s_n}{2}}\)

\(\color{black}{\displaystyle s_{n+2}=\frac{\frac{s_n}{2}}{2}=\frac{2s_n}{2^2}}\)

and then,
\(\color{black}{\displaystyle s_{n+k}=\frac{\frac{s_n}{2}}{2^{k-1}}=\frac{s_n}{2^k}}\)

Answer 9

Probably more useful would be a solution in terms of the initial condition.
\[\begin{align*}
s_{n+1}&=\frac{1}{2}s_n+\frac{1}{6}\\[1ex]
&=\frac{1}{2}\left(\frac{1}{2}s_{n-1}+\frac{1}{6}\right)+\frac{1}{6}\\[1ex]
&=\frac{1}{2^2}s_{n-1}+\frac{1}{6}\left(1+\frac{1}{2}\right)\\[1ex]
&=\frac{1}{2^2}\left(\frac{1}{2}s_{n-2}+\frac{1}{6}\right)+\frac{1}{6}\left(1+\frac{1}{2}\right)\\[1ex]
&=\frac{1}{2^3}s_{n-2}+\frac{1}{6}\left(1+\frac{1}{2}+\frac{1}{2^2}\right)\\[1ex]
&~~\vdots\\[1ex]
&=\frac{1}{2^{k+1}}s_{n-k}+\frac{1}{6}\sum_{i=0}^k\frac{1}{2^i}\\[1ex]
&=\frac{1}{2^{k+1}}s_{n-k}+\frac{2-2^{-k}}{6}
\end{align*}\]Set \(k=n-1\) and you get
\[\begin{align*}
s_{n+1}=\frac{1}{2^n}s_1+\frac{2-2^{-(n-1)}}{6}\implies s_n&=\frac{100}{2^{n-1}}+\frac{2-2^{-(n-2)}}{6}\\[1ex]
s_n&=\frac{1+598\times2^{-n}}{3}
\end{align*}\]Then \(\lim\limits_{n\to\infty}s_n\) is easy to find.

However, it's possible (but probably unlikely), that you mean to find the limit of \(S_n\), where \(S_n\) might denote the \(n\)th partial sum of the sequence. I say unlikely because the partial sums obviously diverge.

Answer 10

Intuitively, ` start at initial value 100 and keep taking the average with 1/3 `.


Doing this, the distance between every new average and 1/3 is divided by 2 at each step. At the limit, the distance is 0.