Let {an} sequence is defined as a1=3, an+1=1/5(an)
converges to zero
a1=3,an+1=15(an)⇒a2=15(a1)=15(3)=35⇒a3=15(a2)=15(35)=325⇒a4=15(a3)=15(325)=3125a_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}a1=3,an+1=51(an)⇒a2=51(a1)=51(3)=53⇒a3=51(a2)=51(53)=253⇒a4=51(a3)=51(253)=1253
Thus, an+1=35na_{n+1}=\dfrac{3}{5^n}an+1=5n3
Now, limn→∞an+1=limn→∞35n\lim_{n\rightarrow \infty}a_{n+1}=\lim_{n\rightarrow \infty}\dfrac{3}{5^n}limn→∞an+1=limn→∞5n3
=3∞=0=\dfrac{3}{\infty} \\=0=∞3=0
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