Sequences & Series (Convergence) MCQ Quiz in తెలుగు - Objective Question with Answer for Sequences & Series (Convergence) - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

Concept:

The necessary condition of convergent series \(\sum u_n\) is  \(\lim_{n\to \infty}u_n=0\).

If \(\lim_{n\to \infty}u_n\neq 0\) then the series \(\sum u_n\) is divergent. 

Explanation:

(1) Let u_n = cos \(\left( \frac{1}{n} \right)\)

\( \Rightarrow \lim _{n \rightarrow \infty} u_n=\lim _{n \rightarrow \infty} \cos \left(\frac{1}{n}\right) \) = cos 0 = 1 ≠ 0

\( \Sigma \cos \left(\frac{1}{n}\right) \) is divergent.

(2) Let \( u_n=\sqrt[7]{\frac{n}{n+1}} \)

∴ \( \lim _{n \rightarrow \infty} u_n=\lim _{n \rightarrow \infty} \sqrt[7]{\frac{n}{n+1}}\) = \(\lim _{n \rightarrow \infty} \sqrt[7]{\frac{n}{1+\frac{1}{n}}} \)

\(=\left(\frac{1}{1+0}\right)^{1 / 7}\) = 1 ≠ 0 

Hence ∑u_n is divergent.

(3) Let \( u_n=n \log \left(\frac{3 n+2}{3 n-2}\right)-1 \)

∴ \(\lim _{n \rightarrow \infty} u_n\) = \(\lim _{n \rightarrow \infty} n \log \left(\frac{3 n+2}{3n -2}\right)\) - 1

 \(=\lim _{n \rightarrow \infty} n \log \left(\frac{1+\frac{2}{3n}}{1 -\frac{2}{3n}}\right)\) - 1

\(=\lim _{n \rightarrow \infty} n \Big[\log(1+\frac{2}{3n}) - \log(1-\frac{2}{3n})\Big]\) - 1

\(=\lim_{n\to\infty}n\Big[2\Big(\frac{2}{3n}+\frac{8}{3\cdot27n^3}+....\Big)\Big]\) - 1

\(=\lim _{n\to \infty}\Big(\frac{4}{3}+\frac{16}{81n^2}+.....\Big)\) - 1

= (\(\frac{4}{3}\) +0 + 0 + .... + 0) - 1

= 1/3 ≠ 0

∴ ∑ u_n is divergent.

Hence option (4) is correct

Concept:

Root Test :

To apply the root test, we need to evaluate

 \(\lim_{n \to \infty} \sqrt[n]{|a_n|}. \)

If this limit is less than 1, the series converges absolutely;

if it is greater than 1, the series diverges;

and if it equals 1, the root test is inconclusive.

Explanation: 
   
Given :  \(|a_n| = \frac{n^{1/3}}{2^n + n}. \)
   

 Apply the root test:
   
 \(\lim_{n \to \infty} \sqrt[n]{|a_n|} \) =  \(\lim_{n \to \infty} \sqrt[n]{\frac{n^{1/3}}{2^n + n}} \) .
   
This expression can be simplified by breaking it into two parts:
   
   \(\sqrt[n]{|a_n|} \)  = \(\frac{\sqrt[n]{n^{1/3}}}{\sqrt[n]{2^n + n}} \) .
   

Evaluate  \(\lim_{n \to \infty} \sqrt[n]{n^{1/3}} \) :

Since  \(n^{1/3} \) grows very slowly compared to exponential terms,
   
 \( \lim_{n \to \infty} \sqrt[n]{n^{1/3}} \) =  \(\lim_{n \to \infty} \left(n^{1/3}\right)^{1/n} \) = 1.
   

Evaluate \( \lim_{n \to \infty} \sqrt[n]{2^n + n} \) :

Since \( 2^n \)  dominates n as \( n \to \infty \), we have
   

 \(\lim_{n \to \infty} \sqrt[n]{2^n + n} \) =  \(\lim_{n \to \infty} \sqrt[n]{2^n} \) =  \(\lim_{n \to \infty} 2 \) = 2.
   

Now,  \(\lim_{n \to \infty} \sqrt[n]{|a_n|} \)  =  \(\frac{1}{2}. \)
   

Since \( \lim_{n \to \infty} \sqrt[n]{|a_n|} \) =  \(\frac{1}{2} \) < 1,

the series \( \sum_{n=1}^{\infty} a_n \)  converges absolutely by the root test.

Hence Option(1) is the correct answer.

Explanation: 

Let

S = \((\frac{1}{2.3}+\frac{1}{2^2.3})+(\frac{1}{2^2.3^2}+\frac{1}{2^3.3^2})+........(\frac{1}{2^a.3^a}+\frac{1}{2^a+1.3^a})\) 

After rearranging the first and second terms of each series, we get, 
\((\frac{1}{2.3}+\frac{1}{2^23^2}+.....\frac{1}{2^a3^a})+\frac{1}{2}(\frac{1}{2.3}+\frac{1}{2^23^2}+....\frac{1}{2^a3^a})+...\)

= \((1+\frac{1}{2})(\frac{1}{2.3}+\frac{1}{2^2.3^2}+...+\frac{1}{2^a3^a}+...)\)

= \(\frac{3}{2}(\frac{1}{2.3}+\frac{1}{2^2.3^2}+...+\frac{1}{2^a3^a}+...)\)

Here we can see that \(\frac{3}{2}(\frac{1}{2.3}+\frac{1}{2^2.3^2}+...+\frac{1}{2^a3^a}+...)\) is geometric series with a= \(\frac{1}{2.3}=\frac{1}{6}\) and r = \(\frac{1}{2.3}=\frac{1}{6}\) < 1.

We know that the sum of infinite geometric series is  \(\frac{a}{1-r}\), r < 1

Hence the sum of the given series is 

S = \(\frac{3}{2}.\frac{a}{1-r} = \frac32. \frac{\frac{1}{6}}{1-\frac{1}{6}}=\frac32.\frac15=\frac3{10}\)

∴ The sum of the series is 3/10.

Explanation:

\(S=\frac{1}{2}-\frac{1}{3 \times 1 !}+\frac{1}{4 \times 2 !}-\frac{1}{5 \times 3 !}+\cdots\)

   = \(\sum_{n=0}^{\infty}\frac{(-1)^n}{(n+2)n!}\)

   = \(\sum_{n=0}^{\infty}\frac{(-1)^n(n+1)}{(n+2)!}\)

   = \(\sum_{n=0}^{\infty}(-1)^n\left[\frac{(n+2)}{(n+2)!}-\frac{1}{(n+2)!}\right]\)

   = \(\sum_{n=0}^{\infty}\frac{(-1)^n}{(n+1)!}-\sum_{n=0}^{\infty}\frac{(-1)^n}{(n+2)!}\)

   = \((1-\frac{1}{2!}+\frac{1}{3!}-\frac{1}{4!}+...)-(\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-...)\)

  = 1 - e^-1- e^-1 (as \((\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-...)=e^{-1}\))

   =1- 2e^-1 = 1 - (2/e)

Option (2) is true

Given: 

\(\frac{4}{a_{n+1}} = \frac{3}{a_n} + \frac{a_n^3}{81}, \quad a_1 = 1.\)

This can be rewritten as:

\(\frac{4}{a_{n+1}} = {\frac{3}{a_n} + \frac{a_n^3}{81}} = \frac{243 + a_n^4} {81 a_n}\) .

⇒ \(a_{n+1}=\frac{324 a_n}{243 + a_n^4} \)

Thus, we have:

\(a_{2} = \frac{324}{243 + 1}\) = \(\frac{324}{244} \) > 1.

Let \(a_{n+1} \)  is an increasing.

Let \(a_{n+1} \) =y and \( a_n \)=x ,then 

y = \(\frac{324 x}{243 + x^4}\) .

To check if y is an increasing function, we calculate \(\frac{dy}{dx} \) :

\(\frac{dy}{dx} \)  =  \(\frac{(243 + x^4)(324) - (324)(4x^3)}{(243 + x^4)^2} \) .

For y to be increasing, we need  \(\frac{dy}{dx} \) > 0, so

\(\frac{dy}{dx} \) =  \(\frac{(243 + x^4)(324) - (324)(4x^3)}{(243 + x^4)^2} \) >0 

⇒ \(243- 3 x^4 >0\)

⇒ \(x^4 < 81\)

⇒ \(x \in (-3,3) \)

Therefore, y is a increasing function for  \(x \in (-3,3) \).

since  \(a_1 \)= 1 and  \(a_n \) = 3 

Limit of  \( a_n\) = [1,3].

Hence Option 3 is the correct answer.

Concept:

Dirichlet’s test: Let ∑ a_n be infinite series such that the series of partial sum S_n = a₁ + a₂ + ... + a_n is bounded. let {b_n} be a monotonically decreasing infinite sequence that converges to zero. Then ∑ a_nb_n converges to some finite value.

Limit comparasion test: Let ∑ an and ∑ bn be two series of positive term and c = \(\displaystyle\lim_{n\to\infty}{a_n\over b_n}\) ≠ 0 and finite then ∑ an and ∑ bn converges and diverges together.

Explanation

(a): \(\displaystyle ∑_{n=1}^{\infty} \frac{\sin (n \pi / 2)}{\sqrt{n}}\)

Here \(∑ \sin ({n\pi\over 2})\) has bounded series of partial sum 

and \(\left\{\frac1{\sqrt n}\right\}\) is a sequence of monotonically decreasing sequence converges to 0

Then by Drichlet's test \(\displaystyle ∑_{n=1}^{\infty} \frac{\sin (n \pi / 2)}{\sqrt{n}}\) is convergent

(a) is convergent

(b): \(\displaystyle ∑_{n=1}^{\infty} \log \left(1+\frac{1}{n^2}\right).\)

Let a_n = \( \log \left(1+\frac{1}{n^2}\right)\) and b_n = \(\frac1{n^2}\)

Then \(\displaystyle\lim_{n\to\infty}{a_n\over b_n}\) = \(\displaystyle\lim_{n\to\infty}{ \log \left(1+\frac{1}{n^2}\right)\over \frac1{n^2}}\) (0/0 form)

                      = \(\displaystyle\lim_{n\to\infty}{\frac1{1+\frac{1}{n^2}}.(-{\frac2{n^3}})\over -{\frac2{n^3}}}\) (L'hospital rule)

                   = 1 ≠ 0

So, by limit comparasion test, ∑ an and ∑ bn converges and diverges together.

Now, ∑ bn = ∑ \(\frac1{n^2}\) is convergent series by p-series test.

Hence \(\displaystyle ∑_{n=1}^{\infty} \log \left(1+\frac{1}{n^2}\right)\) is convergent

(b) is convergent

(3) is correct

Concept:

(i) Limit Comparison Test : 

If, (a) a_n ≥ 0 and b_n ≥ 0 , ∀ n

(b) \(lim_{n \rightarrow \infty} \frac{a_n}{b_n}=c\), Where c is finite positive constant.

Then, the series \(\sum_{n=1 }^{\infty} {\ a_n}\) and \(\sum_{n=1 }^{\infty} {\ b_n}\) behave alike 

(ii) p- series test :

If p > 1. Then the series \(\sum_{n=1 }^{\infty} {\frac{1}{n^p}}\) converges.

If p ≤ 1. Then the series \(\sum_{n=1 }^{\infty} {\frac{1}{n^p}}\) diverges.

Explanation:

(a) Let a_n=sin(\(\frac{1}{n}\)) and b_n= \(\frac{1}{n}\)

Now, \(lim_{n \rightarrow \infty} \frac{sin(\frac{1}{n})}{\frac{1}{n}}=1\) \(\neq\)0(finite) 

By using limit Comparison test

 \(\sum_{n=1 }^{\infty}sin {\frac{1}{n}}\) and \(\sum_{n=1 }^{\infty} {\frac{1}{n}}\) behave alike 

Now, By using p - series test \(\sum_{n=1 }^{\infty} {\frac{1}{n}}\) diverges 

⇒ \(\sum_{n=1 }^{\infty}sin {\frac{1}{n}}\) is also diverges.

(b) \(\frac{1}{3^2} \frac{2}{4^2}+\frac{3}{5^2} \cdot \frac{4}{6^2} \)+\(\frac{5}{7^2} \frac{6}{8^2}+\ldots\) = \(\sum_{n=1 }^{\infty} {\frac{(2n-1)2n}{(2n+1)^2(2n+2)^2}}\)

Let a_n= \( {\frac{(2n-1)2n}{(2n+1)^2(2n+2)^2}}\) and bn= \(\frac{1}{n^2}\)

\(lim_{n \rightarrow \infty} \frac{(2n-1).2n.n^2}{(2n+1)^2.(2n+2)^2}=\frac{1}{4}\)​\(\neq\)0(finite)​​

By using limit Comparison test

 \(\sum_{n=1 }^{\infty} {\frac{(2n-1)2n}{(2n+1)^2(2n+2)^2}}\) and \(\sum_{n=1 }^{\infty} {\frac{1}{n^2}}\) behave alike 

Now, By using p - series test \(\sum_{n=1 }^{\infty} {\frac{1}{n^2}}\) converges. 

⇒ \(\sum_{n=1 }^{\infty} {\frac{(2n-1)2n}{(2n+1)^2(2n+2)^2}}\) is also converges.

Hence, (1) option is true

Concept:

Integral test : This test is used for series of the form \(\sum_{n=1}^{\infty} f(n)\) where f(x)  is a continuous, positive, and decreasing function for x ≥ 1 

The series converges if and only if the corresponding improper integral \(\int_{1}^{\infty}f(x) dx\) converges.

Explanation:

\(\lim_{n\rightarrow\infty}\left(\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2 n}\right)\)

S_n = \(\sum_{n=1}^{\infty} \frac{1}{n+k}\)

\(lim_{n\rightarrow \infty}\) S_n = \(lim_{n\rightarrow \infty}\) \(\frac{1}{n}\)\(\sum_{n=1}^{\infty} \frac{1}{1+\frac{k}{n}}\) 

= \(\int_{0}^{1}\frac{1}{1+x}\rm dx\)

= log_e\(\)\((1+x)|_{0}^{1}\)

= log_e2

Explanation - If every subsequence of a sequence is convergent then sequence is convergent 

Option 1:-  Let 

\(x_n =\begin{cases} 4, \text{n is odd} \\ 0, \text{ n is even} \end{cases}\)  and  \(x_{2n} = 4 ; x_{2n+1} = 0 \)  

Here subsequence is convergent but sequence oscillate between 0 and 4 . 

Therefore , Option 1 is False.

Option 2 :- Let x_n = (-1)ⁿ then  {(-1)nxn} = {1} which is converges but {xn} does not converges

\(\therefore\) Option 2 is false.

Option 3:- By Cauchy Second Theorem on limit

If \(\lim_{n\to\infty} {\frac{x_{n+1}}{x_{n}}}\) exist then \(\lim _{n\to \infty}(x_n)^{\frac{1}{n}} \) also exist

and so bounded. 

Option 3) is True.

Option 4 :- Let \(x_n =\begin{cases} 4, \text{n is prime} \\ 0, \text{elsewhere} \end{cases}\)

It is unbounded sequence but  here one subsequence is 4 which is bounded  so, Option 4 is False. 

Concept:

For |a| < 1, the expansion of ln(1 + a) is given by 

\(\ln(1+a) = a-\frac{a^2}{2}+\frac{a^3}{3}-\frac{a^4}{4}+\cdots,= \sum_{x = 1}^{\infty} (-1)^{x+1} \frac{a^x}{x}\)

Calculation:

We know that,

\(\ln(1+x) = x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots,\)

Substituting x by -x, we get:

\(\ln(1-a) = -a-\frac{a^2}{2}-\frac{a^3}{3}-\frac{a^4}{4}+\cdots, = -\sum_{x = 1}^{\infty} \frac{a^x}{x}\)

Now, \(S_1=\frac{1}{3} - \frac{1}{2} \times \frac{1}{{{3^2}}} + \frac{1}{3} \times \frac{1}{{{3^3}}} - \frac{1}{4} \times \frac{1}{{{3^4}}} + ...\)

⇒ \(S_1=\sum_{x=1}^{\infty}(-1)^{x+1}\frac{1}{x}\cdot\frac{1}{3^x}\)

⇒ \(S_1=\sum_{x=1}^{\infty}(-1)^{x+1}\frac{\left(\frac{1}{3^x}\right)}{x}\)

⇒ \(S_1=\sum_{x=1}^{\infty}(-1)^{x+1}\frac{\left(\frac{1}{3}\right)^x}{x}\)

⇒ \(S_1=\ln \left(1+\frac{1}{3}\right)\)

⇒ \(S_1=\ln \left(\frac{4}{3}\right)\)

Now, \(S_2=\frac{1}{4} + \frac{1}{2} \times \frac{1}{{{4^2}}} + \frac{1}{3} \times \frac{1}{{{4^3}}} + \frac{1}{4} \times \frac{1}{{{4^4}}} + ...\)

⇒ \(S_2=\sum_{x=1}^{\infty}\frac{1}{x}\cdot\frac{1}{4^x}\)

⇒ \(S_2=\sum_{x=1}^{\infty}\frac{\left(\frac{1}{4^x}\right)}{x}\)

⇒ \(S_2=\sum_{x=1}^{\infty}\frac{\left(\frac{1}{4}\right)^x}{x}\)

⇒ \(S_2=-\ln \left(1-\frac{1}{4}\right)\)

⇒ \(S_2=-\ln \left(\frac{3}{4}\right)\)

⇒ \(S_2=\ln \left(\frac{4}{3}\right)\) [∵ - ln x = ln (\(\frac{1}{x}\))]

∴ \(S_1=S_2\)