Question

Suppose that (x_n) is a sequence of positive terms which satisfies x_(n+1) <= (1/2)x_n (*) for all n >= 1. Use induction to prove that x_(n+1) <=(1/2^(n-1))x_1
for all n>=1. Deduce that x_n --> 0 as n -->infinity.

Answer 1

What is (*) ?

Answer 2

yes, induction will do it
i would start with the case \(k=2\) as the base case, see that \(x_2\leq \frac{1}{2}x_1\) is true by assumption
then assume true for all \(n\leq k\) and show it is true for \(k+1\)

Answer 3

induction hypothesis is that \(x_k\leq \left(\frac{1}{2}\right)^{k-1}x_{k-1}\)

Answer 4

then \(x_{k+1}\leq \frac{1}{2}x_k\) by assumption, and by induction \(x_k\leq \left(\frac{1}{2}\right)^{k-1}x_1\) and so
\[x_{k+1}\leq \frac{1}{2}x_k\leq \frac{1}{2}\left(\frac{1}{2}\right)^{k-1}x_1=\left(\frac{1}{2}\right)^{k}x_1\]

Answer 5

crap i wrote induction hypothesis wrong, my fault!!

Answer 6

induction hypothesis is that \(x_k\leq \left(\frac{1}{2}\right)^{k-1}x_1\)

Answer 7

sorry about that, probably caused some confusion

Answer 8

yh should be \[\frac{ 1 }{ 2^{k-1} }\] right

Answer 9

i guess that was the part you meant

Answer 10

then \[x _{k+1}\le \frac{ 1 }{ 2 }x _{k}\] by assumption, and by induction \[x _{k}\le \frac{ 1 }{ 2^{k-1} }x _{k1}\] and so \[x _{k+1}\le \frac{ 1 }{ 2 }x _{k}\le \frac{ 1 }{ 2} \frac{ 1 }{ 2^{k-1}}x_{1}= \frac{ 1 }{ 2^{k}}x_{1}\]

is this right, im not sure.?