Let $\{v_n\}$ be a sequence defined by $v_1=1$ and $v_{n+1}=\sqrt{v_n^2 +(1/5)^n}$, for $n\ge1$. Then find limit of the sequence $\{v_n\}$. I have found that the given sequence is monotone increasing but failed to find the limit. Please help me to solve it.
Asked
Active
Viewed 199 times
1 Answers
11
If I have read your question aright, then $$v_{n+1}^2=v_n^2+\frac1{5^n}.$$ So $$\lim_{n\to\infty}v_n^2=1+\sum_{n=1}^\infty\frac1{5^n}.$$ You can sum this geometric series and take its square root to get the limit you seek.
Angina Seng
- 158,341
-
Thank you. Good idea. @Lord Shark the Unknown – Iamdark Jun 11 '17 at 06:20
-
5@Iamdark If Lord Shark's answer answers your question, you should accept it. – AlgorithmsX Jun 11 '17 at 07:31
-
@AlgorithmsX It is a good idea to wait some time. Maybe other user provide answers that pleases Iamdark more. Also it gives other user the chance to check the answer. At least two or three days are absolutely reasonable.reasonable. – miracle173 Jun 11 '17 at 12:08
-
@miracle173 Fair enough. – AlgorithmsX Jun 11 '17 at 14:45