Question

Let {an} sequence is defined as a1=3, an+1=1/5(an)

converges to zero. Prove that

Answer 1

Solution:

$a_1=3, a_{n+1}=\dfrac15(a_n) \\ \Rightarrow a_{2}=\dfrac15(a_1)= \dfrac15(3)=\dfrac35 \\ \Rightarrow a_{3}=\dfrac15(a_2)= \dfrac15(\dfrac35)=\dfrac3{25} \\ \Rightarrow a_{4}=\dfrac15(a_3)= \dfrac15(\dfrac3{25})=\dfrac3{125}$

Thus, $a_{n+1}=\dfrac{3}{5^n}$

Now, $\lim_{n\rightarrow \infty}a_{n+1}=\lim_{n\rightarrow \infty}\dfrac{3}{5^n}$

$=\dfrac{3}{\infty} \\=0$