Mathematics: CSIR-NET Previous year Solved questions on Sequence.

1. Let $a_n=\sin(\pi/n)$. For the sequence $a_1, a_2, \cdots$ the supremum is

• $0$ and it is attained
• $0$ and it is not attained
• $1$ and it is attained
• $1$ and it is not attained

 Solution: This is sequence is {sin$\pi$, sin$\pi/2$, sin$\pi/3$$\dots$}

Clearly, Supremum is 1. So, 3rd option is correct.

2.Which of the following is/are correct?

• $n \log(1+ \frac{1}{n+1}) \rightarrow 1$ as $n \rightarrow \infty$
• $(n+1) \log(1+ \frac{1}{n+1}) \rightarrow 1$ as $n \rightarrow \infty$
• $n^2 \log(1+ \frac{1}{n}) \rightarrow 1$ as $n \rightarrow \infty$
• $n \log(1+ \frac{1}{n^2}) \rightarrow 1$ as $n \rightarrow \infty$

Solution:

$\lim_{n\to \infty}\frac{1+\log n}{n}=1$

Option $1$ is correct:

$n \log(1+ \frac{1}{n+1})$ =$n \log(1+\frac{1}{n+1})=(n+1)\log(1+ \frac{1}{n+1})-\log(1+\frac{1}{n+1})$ So $\lim_{n\to \infty}\frac{\log(1+\frac{1}{n+1})}{n+1}-\lim_{n\to \infty}\log(1+\frac{1}{n+1})$=$\lim_{\frac{1}{n+1}\to 0}\frac{\log(1+\frac{1}{n+1})}{n+1}-\lim_{\frac{1}{n+1}\to 0}\log(1+\frac{1}{n+1})=1+0=1$

Option $2$ is correct:

$(n+1) \log(1+ \frac{1}{n+1})=\frac{\log(1+ \frac{1}{n+1})}{\frac{1}{n+1}}$ taking limit :

$\lim_{\frac{1}{n+1} \to 0}\frac{\log(1+ \frac{1}{n+1})}{\frac{1}{n+1}}=1$

Option $3$:

$n^2 \log(1+ \frac{1}{n})=n\times \frac{\log(1+ \frac{1}{n})}{1/n}$ taking limit:

we get: $\lim_{n\to \infty} n \times \lim_{\frac{1}{n}\to 0}n^2 \log(1+ \frac{1}{n})$=$\infty \times 1=\infty$

Option $4$:

$\lim_{n\to \infty} n \log(1+ \frac{1}{n^2})=\lim_{n\to \infty} \frac{n^2}{n} \log(1+ \frac{1}{n^2})=\lim_{n\to \infty}\frac{1}{n}\times\frac{\log(1+\frac{1}{n^2})}{\frac{1}{n^2}}=0$.

So, $1$ and $4$ are correct options.

3.If $\{x_n \}$ and $\{y_n \}$ are sequence of real numbers, which of the following is/are true?

• $\limsup (x_n+y_n) \leq \limsup x_n + \limsup y_n$
• $\limsup (x_n+y_n) \geq \limsup x_n + \limsup y_n$
• $\liminf (x_n+y_n) \leq \liminf x_n + \liminf y_n$
• $\liminf (x_n+y_n) \geq \liminf x_n + \liminf y_n$

Solution:

Given two sequences, $\;\displaystyle \{a_n\}_{n \in \Bbb N},\;\,\{b_n\}_{n \in \Bbb N},\;$

and given the definition of the supremum of a sequence, we can see that for every $k\geq n$, $(a_k + b_k) \;\; \leq \;\;\sup_{k\geq n} a_k + \sup_{k\geq n} b_k\,.$

So , $Sup_{k\geq n}(a_k + b_k) \;\; \leq \;\;\sup_{k\geq n} a_k + \sup_{k\geq n} b_k\,.$

 Taking limit both sides we get $\limsup (x_n+y_n) \leq \limsup x_n + \limsup y_n$

Similarly, we can prove that,

 $\liminf (x_n+y_n) \geq \liminf x_n + \liminf y_n$

7.$\lim_{n \rightarrow \infty} \frac{1}{\sqrt{n}}(\frac{1}{\sqrt{1} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{5}} + \cdots + \frac{1}{\sqrt{2n - 1} + \sqrt{2n + 1}})$ equals

• $\sqrt{2}$
• $\frac{1}{\sqrt{2}}$
• $\sqrt{2} + 1$
• $\frac{1}{\sqrt{2} + 1 }$

Solution: $\lim_{n \rightarrow \infty} \frac{1}{\sqrt{n}}(\frac{1}{\sqrt{1} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{5}} + \cdots + \frac{1}{\sqrt{2n - 1} + \sqrt{2n + 1}})$

=$\lim_{n \rightarrow \infty} \frac{1}{\sqrt{n}}(\frac{\sqrt{1}-{\sqrt{3}}}{-2} + \frac{\sqrt{3}-\sqrt{5}}{-2} + \cdots + \frac{\sqrt{2n - 1}-\sqrt{2n + 1}}{-2})$

=$lim_{n\to \infty}$ $\frac{1}{\sqrt{n}}$ $(\sqrt{1}-\sqrt{2n+1})$

=$\frac{\sqrt{2}}{2}$=$\frac{1}{\sqrt{2}}$.

8.Let $p(x)$ be a polynomial in the real variable $x$ of degree $5$. Then $\lim_{n\rightarrow \infty} \frac{p(n)}{2^n}$ is

•  $5$
• $1$
• $0$
• $\infty$

Solution: Let $P(x)=a_5 x^5+a_4 x^4+\dot a_0$ and using binomial expansion , $2^x=(1+1)^x=1+x+\frac{x(x-1)}{2!}+...$, Dividing numerator and denominator by $x^5$, and taking limit $x\to \infty$ we get $lim_{x\to \infty} \frac{a_5+a_4/x+...+a_0/x^5}{1+1/x^4+x(x+1)/(2\times x^5)+...+(x(x-1)(x-2)(x-3)(x-4))/5!x^5+x(x-1)(x-2)(x-3)(x-4)(x-5))/6!x^5+....)}$

So as denominator tends to $\infty$ and numerator tends to $a_5$ (A constant value) So limit tends to zero.

9..The limit

$\lim_{x \rightarrow 0} \frac{1}{x} \int_x^{2x}e^{-t^2}~dt$

1. does not exist.
2. is inifinite
3. exists and equals 1 .
4. exists and equals 0 .

Solution: 

$\lim_{x\rightarrow 0}\frac{1}{x}\int_x^{2x}e^{-t^2}~dt$=$\lim_{x\rightarrow 0}\frac{1}{x}\int_x^{2x}\left[ 1-\frac{t^2}{1!}+\frac{t^4}{2!}-\cdots \right]~dt$

=$\lim_{x\rightarrow 0} \frac{1}{x}\left[ t-\frac{t^3}{3\cdot 1!}+\frac{t^4}{4\cdot 2!}-\cdots \right]_x^{2x}$

=$\lim_{x \rightarrow 0} \frac{1}{x} \left[ x-\frac{(2x)^3-x^3}{3\cdot 1!}+\frac{(2x)^4-x^4}{4\cdot 2!} - \cdots \right]$

= $\lim_{x \rightarrow 0} \frac{1}{x} \left[x-\frac{7x^3}{3\cdot 1!}+\frac{15x^4}{4\cdot 2!} - \cdots \right]$

=$\lim_{x \rightarrow 0}\left[1-\frac{7x^2}{3\cdot 1!}+\frac{15x^3}{4\cdot 2!}-\cdots \right]=1$