Sequences and Series MCQ Quiz in मल्याळम - Objective Question with Answer for Sequences and Series - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Given:

\(\sqrt{\frac{1}{2}} + \sqrt{\frac{2}{3}} + \sqrt{\frac{3}{4}} + \sqrt{\frac{4}{5}} + .....\)

Concept used: 
The necessary condition of series to be convergent is \(\underset{n \rightarrow \infty}{L t} {u_n}= 0\);

Limit comparison test:

if un and vn are positive terms sequence such that Ltn→∞unvn=c" id="MathJax-Element-6-Frame" role="presentation" style="display: inline; position: relative;" tabindex="0">Ltn→∞unvn=cLtn→∞unvn=c

 where c > 0 and c is finite then Both un and vn converge and diverge together

P- series test: 

∑ \(\frac{1}{n^p}\)1np" id="MathJax-Element-25-Frame" role="presentation" style="display: inline; position: relative;" tabindex="0">1np1np is convergent for p > 1 and divergent for p ≤  1

u_n = \(\underset{n \rightarrow \infty}{L t} \sqrt{\frac{n}{n + 1 }}=1\) 

⇒ necessary condition fails 

Now, by limit comparison test 

Take 

\(\underset{n \rightarrow \infty}{L t} \frac{u_n}{v_n}= \underset{n \rightarrow \infty}{L t} \sqrt{\frac{1}{1+\frac{1}{n}}}=1\) (Finite)

∑v_n is divergent by p-series test. (p = 0 < 1)

∴ By limit comparison test, ∑u_n is divergent

∴ option 2 is correct 

Concept:

\(log(1-x)\rm = \left( - x - \frac{x^2}{2 } - \frac{x^3}{3 } - \frac{x^4}{4} ... \infty\right)\)

\(log (1+x)= \left( x - \frac{x^2}{2 } +\frac{x^3}{3 } - \frac{x^4}{4} ... \infty\right)\)

Calculation:

We know that,

\(\rm \log_e \left( \frac{1-x}{1 + x} \right) = \log_e ( 1 - x) - \log_e( 1 + x)\)

\(\rm = \left( - x - \frac{x^2}{2 } - \frac{x^3}{3 } - \frac{x^4}{4} ... \infty\right)- \left( x - \frac{x^2}{2 } +\frac{x^3}{3 } - \frac{x^4}{4} ... \right)\)

\(\rm = -2x - 2 \frac{x^3}{3} - 2 \frac{x^5}{5} ....\)

\(\rm \log_e \left( \frac{1-x}{1+x} \right) = -2 \left[ x + \frac{x^3}{3} + \frac{x^5}{5} + ... \right] \)

\(- \frac{1}{2} \log_e \left( \frac{1-x}{1+x} \right) = x + \frac{x^3}{3} + \frac{x^5}{5} +....\)

Therefore,

\(\rm \frac{m -n}{m +n} + \frac{1}{3} \left( \frac{m -n}{m+n} \right)^3 + \frac{1}{5} \left( \frac{m -n}{m+n} \right)^5 + .... \infty\)

\(= \frac{-1}{2} \log_e \frac{1 - \left( \frac{m -n}{m + n} \right)}{1+ \left( \frac{m -n}{m + n} \right)} \)

\(\rm = \frac{-1}{2} \log_e \left( \frac{m + n-(m - n)}{m + n + m - n} \right)\)

\(\rm = \frac{-1}{2} \log_e \left( \frac{2n}{2m} \right) = \frac{-1}{2} \log_e\left( \frac{m}{n} \right)\)

Let the given expansion be f(n)

f(n) = (1 + a1)(1 + a2)... (1 + an)

Also given, a1 = a2 = a3 = ... = an = 1

Consider for n = 2

f(2) = (1 + a1)(1 + a2)

f(2) = (1 + 1)(1 + 1) = 2²

Consider for n = 5

f(5) = (1 + a1)(1 + a2)(1 + a3)(1 + a4)(1 + a5)

f(5) = (1 + 1)(1 + 1)(1 + 1)(1 + 1)(1 + 1) = 25

Similary for n times, it is given as

f(n) = (1 + 1)(1 + 1) ...... (1 + 1) = 2ⁿ

(1 + a1)(1 + a2)... (1 + an) = 2n

Concept Used:-

If the summation from 0 to n such that \(\rm\displaystyle\sum_{r = 0}^{ n}\) ar xr is given, then the expanded form of it can be written as,

 ⇒ \(\rm\displaystyle\sum_{r = 0}^{ n}\) ar xr = a₀+a₁ x+a₂ x²+a₃ x³+a₄ x⁴+..........+a_nxⁿ

Explanation:-

Given,

(1+ x + x2)n = \(\rm\displaystyle\sum_{r = 0}^{2 n}\) ar xr, 

On the expanding right-hand side, we get,

\(\left(1+x+x^2\right)^n=a_0+a_1 x+a_2 x^2+a_3 x^3+a_4 x^4+\cdots+a_{2 n} x^{2 n}\)

Differentiate it with respect to x,

\(n\left(1+x+x^2\right)^{n-1}(1+2 x)=a_1+2 a_2 x+3 a_3 x^2+4 a_4 x^3+\cdots+2 n a_n x^{2 x-1}\)Now put x = -1 in the above equation,

\(\Rightarrow n(1-1+1)^{n-1}(1-2)=a_1-2 a_2+3 a_3-4 a_4+5a_5 \cdots-2 n a_{2n}\\ \Rightarrow n^1(1)^{n-1}(-1)=a_1-2 a_2+3 a_3-4 a_4+5a_5 \cdots-2 n a_{2n}\\ \Rightarrow -n=a_1-2 a_2+3 a_3-4 a_4+5a_5 \cdots-2 n a_{2n}\)

So, the value of a1 − 2a2 + 3a3 − …. −2n a2n is -n.

Hence, the correct option is 3.

Given:

The Sequence mn where m, n ∈ N 

Concept used:

The Sequence is bounded if for all the Terms of the sequence there exists M and m such that m < a_i < M, where ai, are the terms in the Sequence

if no such m and M exists then Sequence is Unbounded 

if revolving around a number than  it is Oscillating.

Calculations:

The Terms of the Sequence mⁿ are as 0, 1, 2, 4, 8,.......,3, 9, 27,........, 4, 16, 64,........ as so on.

So there doesn't exist M such that all the Terms of the Sequence are less than that M and m = 0 here'

∴ The sequence is not bounded i.e Unbounded.

Concept used:

Ratio test:

L = \(\mathop {Lt}\limits_{n \to \infty } \frac{{{u_{n + 1}}}}{{{u_n}}} \)

P - Series test: 

\({u_n} = \frac{{(n + 1){x^n}}}{{{n^3}}}; {u_{n + 1}}\frac{{(n + 2){x^{n + 1}}}}{{{{(n + 1)}^3}}}\)

\(\frac{{{u_{n + 1}}}}{{{u_n}}} = \frac{{n + 2}}{{{{(n + 1)}^3}}}.{x^{n + 1}}.\frac{{{n^3}}}{{(n + 1){x^n}}} = \left( {\frac{{n + 2}}{{n + 1}}} \right){\left( {\frac{n}{{n + 1}}} \right)^3}.x\)

\(\mathop {Lt}\limits_{n \to \infty } \frac{{{u_{n + 1}}}}{{{u_n}}} = \mathop {Lt}\limits_{n \to \infty } \left( {\frac{{1 + \frac{2}{n}}}{{1 + \frac{1}{n}}}} \right)\frac{1}{{{{\left( {1 + \frac{1}{n}} \right)}^3}}}.x = x \)

∴ By ratio test, ∑un converges when x < 1 and diverges when x > 1.

When x = 1, \({u_n} = \frac{{n + 1}}{{{n^3}}}\) 

Take \({v_n} = \frac{1}{{{n^2}}}\) ; By Limit comparison test ∑u_n is convergent 

∴ ∑u_n is convergent if x ≤  1 and divergent if x > 1.

Given:

\(∑_{n = 1}^\infty \sin \left( \frac{1}{n} \right)\)

Concept used:

Limit Comparision test:

if an and bn are two positive series such that \(\underset{n \rightarrow \infty}{L t} \frac{u_{n}}{v_{n}}= c\)Ltn→∞anbn=c" id="MathJax-Element-24-Frame" role="presentation" style="display: inline; position: relative;" tabindex="0"> 

where c > 0 and finite then, either Both series converges or diverges together

P - series test:

 ∑ \(\frac{1}{n^p}\)is convergent for p > 1 and divergent for p ≤  1

Calculations:

\(∑_{n = 1}^\infty \sin \left( \frac{1}{n} \right)\)

take v_{n =} 

\(\underset{n \rightarrow \infty}{L t} \frac{u_{n}}{v_{n}}= \underset{n \rightarrow \infty}{L t} \frac{\sin \left( \frac{1}{n} \right)}{\left( \frac{1}{n} \right)} = \underset{t \rightarrow 0}{L t} \frac{\sin t}{t} \) (where t = 1/n) = 1

∴ ∑u_n, ∑v_n both converge or diverge. But ∑v_n = \(∑ \frac{1}{n}\) is divergent

(p-series test, p = 1);

∴ ∑u_n is divergent.

Concept:

By ratio test, if \(\mathop {\lim }\limits_{n \to \infty } \frac{{{u_n}}}{{{u_{n + 1}}}} = \lambda ,\)

Then the series converges for λ > 1 and diverges for λ < 1 and fails for λ =1;

By Raabe’s test, if \(\mathop {\lim }\limits_{n \to \infty } n\left[ {\frac{{{u_n}}}{{{u_{n + 1}}}} - 1} \right] = k,\)

Then the series converges for k > 1 and diverges for k < 1, but the test fails for k = 1.

Calculation:

From the given series,

\(\frac{{{u_n}}}{{{u_{n + 1}}}} = {\left( {\frac{{n!}}{{\left( {n + 1} \right)!}}} \right)^2}.\frac{{\left( {2\left( {n + 1} \right)} \right)!}}{{\left( {2n} \right)!}}.\frac{{{x^{2n}}}}{{{x^{2\left( {n + 1} \right)}}}}\)

\( \Rightarrow \frac{{{u_n}}}{{{u_{n + 1}}}} = \frac{{\left( {2n + 1} \right)\left( {2n + 2} \right)}}{{{{\left( {n + 1} \right)}^2}}}\frac{1}{{{x^2}}} = \frac{{2\left( {2n + 1} \right)}}{{\left( {n + 1} \right)}}\frac{1}{{{x^2}}}\)

Taking limit,

\(\mathop {\lim }\limits_{n \to \infty } \frac{{{u_n}}}{{{u_{n + 1}}}} = \mathop {\lim }\limits_{n \to \infty } \frac{{2\left( {2 + \frac{1}{n}} \right)}}{{1 + \frac{1}{n}}}\frac{1}{{{x^2}}} = \frac{4}{{{x^2}}}\)

Thus by ratio test, series converges for x² < 4 and diverges for x² > 4, but fails for x² = 4;

When x² = 4,

\(n\left[ {\frac{{{u_n}}}{{{u_{n + 1}}}} - 1} \right] = n\left[ {\frac{{2n + 1}}{{2\left( {n + 1} \right)}} - 1} \right] = - \frac{n}{{2n + 2}}\)

\(\mathop {\lim }\limits_{n \to \infty } n\left[ {\frac{{{u_n}}}{{{u_{n + 1}}}} - 1} \right] = - \frac{1}{2} < 1\)

Thus by Raabe's test, the series diverges.

Hence the given series

converges for x² < 4 → -2 < x < 2 and

diverges for x² ≥ 4 → x ≤ -2 and x ≥ 2

Concept used:

Ratio test:

L = \(\mathop {Lt}\limits_{n \to \infty } \frac{{{u_{n + 1}}}}{{{u_n}}} \)

L > 1 the series is Divergent neither convergent or divergent 

L < 1 the series is Convergent 

L = 1 test fails Neither Convergent nor Divergent 

Calculations:

u_n = n^1-n ; u_{n + 1} = (n + 1)^-n ;

\(\frac{{{u_{n + 1}}}}{{{u_n}}} = \frac{{{{(n + 1)}^{ - n}}}}{{{n^{1 - n}}}} = \frac{{{n^n}}}{{n{{(n + 1)}^n}}} = \frac{1}{n}{\left( {\frac{n}{{n + 1}}} \right)^n} \)

\(\mathop {Lt}\limits_{n \to \infty } \frac{{{u_{n + 1}}}}{{{u_n}}} = \mathop {Lt}\limits_{n \to \infty } \frac{1}{n}.{\left( {\frac{1}{{1 + \frac{1}{n}}}} \right)^n} = 0.\frac{1}{e} = 0 < 1 \)

∴ By ratio test ∑u_n, is convergent

Given:

\(\rm u_n=\left(\frac{1}{n}\right).\sin\left(\frac{1}{n}\right)\)

Concept used:

Limit Comparision test:

if an and bn are two positive series such that \(\underset{n \rightarrow \infty}{L t} \frac{a_n}{b_n} = c \) 

where c > 0 and finite then, either Both series converges or diverges together

P - Series test:

∑ \(\frac{1}{n^p }\)is convergent for p > 1 and divergent for p ≤  1

Calculations:

Let v_n  = \(\frac{1}{n^2 }\) , so that Σv_n is convergent by p-series test.

\(\rm \displaystyle Lt_{n\rightarrow \infty}\left(\frac{u_n}{v_n}\right)=\rm \displaystyle Lt_{n\rightarrow \infty}\frac{\sin \left(\frac{1}{n}\right)}{\left(\frac{1}{n}\right)}\)

\(=\rm \displaystyle Lt_{n\rightarrow 0}\left(\frac{\sin t}{t}\right)\)

where t = 1/n, Thus \(\rm \displaystyle Lt_{n\rightarrow \infty}\left(\frac{u_n}{v_n}\right)=1\ne0\)

∴ By Limit comparison test Σu_n is convergent.