ISI2017-MMA-14

Question

Let $(v_n)$ be a sequence defined by $v_1 = 1$ and $v_{n+1} = \sqrt{v_n^2 +\left(\dfrac{1}{5}\right)^n}$ for $n\geq1$. Then $\displaystyle{\lim_{n \rightarrow \infty}v_n}$ is

• A. $\sqrt{5/3}$
• B. $\sqrt{5/4}$
• C. $1$
• D. $\text{non existent}$

Comments

Not able to understand

Answer 1

$V_{n+1} = \sqrt{V_{n}^2 + (1/5)^n}$

on  squaring both side, we get

$V_{n+1}^2 - V_{n}^2 = (1/5)^n$

$V_{2}^2 - V_{1}^2 = (1/5)$

$V_{3}^2 - V_{2}^2 =  (1/5)^2$

$V_{4}^2 - V_{3}^2 =  (1/5)^3$

$V_{5}^2 - V_{4}^2 =  (1/5)^4$

$\vdots$

$V_{n}^2 - V_{n-1}^2 = (1/5)^n-1$

On adding all we get,

$V_{n}^2 - V_{1}^2 = ((1/5) + (1/5)^2 + (1/5)^3 + (1/5)^4 + (1/5)^5 +\dots + (1/5)^{n-1}$

$\quad = ( 1 - (1/5)^{n-1}) / 4$

$\quad =  1/4 - (1/5)^{n-1}/4 $

$\implies V_{n}^2 =  V_{1}^2 +  1/4 - (1/5)^{n-1}/4 $

$\quad \quad  =  1+ 1/4 - (1/5)^{n-1}/4 $

$\quad \quad  = 5/4 - (1/5)^{n-1}/4 $

on taking limit, we get

$V_{n} = \sqrt{5/4}$

$$(v)\lim_{x\to \infty}a^x = \begin{cases}0, &0\leq a < 1 \\1, \quad &a= 1\\\infty, &a>1\\\text{does not exist}, & a < 0\end{cases}$$

Correct Answer: $B$

Comments

Can you explain the part of taking limit and afterwards? What is this procedure?

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$\lim_{n \rightarrow \infty} \frac{1}{5^{n-1}}=\lim_{n \rightarrow \infty} \frac{5}{5^n} = 0$ because as $n$ is increasing, $5^n$ is getting bigger and bigger and so,  $5^n$ approaches infinity when $n$ approaches infinity and so, $\frac{1}{5^n}$ approaches zero. It means $\lim_{n \rightarrow \infty} \frac{5}{5^{n}}=0$ which is used here.

You can write given recurrence as:

$v_{n}^2 = v_{n-1}^2 +\frac{1}{5^{n-1}}$ (replace $n$ with $n-1$ )

$v_{n}^2 = v_{n-2}^2 +\frac{1}{5^{n-2}}+\frac{1}{5^{n-1}}$

When expand right hand side like this,at some instance, it becomes,

$v_{n}^2 =v_1^2+\frac{1}{5^1}+ …….+\frac{1}{5^{n-2}}+\frac{1}{5^{n-1}}$

Put $v_1=1$

$v_{n}^2 =1+\frac{1}{5^1}+ …….+\frac{1}{5^{n-2}}+\frac{1}{5^{n-1}}$

Right hand side is sum of GP Series with $n$ terms, So,

$v_{n}^2 =\frac{1\times (1-\frac{1}{5^n})}{1 – \frac{1}{5}}$

$v_{n}^2 =\frac{5}{4}(1-\frac{1}{5^n})$

Since, $\lim_{n \rightarrow \infty} \frac{1}{5^{n}}=0$

So, $\lim_{n \rightarrow \infty}v_{n}^2 = \lim_{n \rightarrow \infty} \frac{5}{4}(1-\frac{1}{5^n}) = \frac{5}{4}(1-0) = \frac{5}{4}$

So, $\lim_{n \rightarrow \infty}v_{n}= \sqrt{\frac{5}{4}}$

Answer 2

Vn=root(1+(1/5)+(1/5)^2+.......(1/5)^n-1)

lim n->infinite Vn=root((5/4)-(1/(4*5^(n-1))))  =root(5/4)

answer (B)