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

solution- 

we have a sequence $\{y_n{\}}_{n=1}^{\infty }$

If $\{y_n{\}}_{n=1}^{\infty }$ tends to 1 as n tends to infinity then yn → 1 as n → ∞ 

Therefore, Correct answer is Option 3).

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 $\underset{n\to \mathrm{\infty }}{Lt}{u}_n=0$;

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=c$\underset{n\to \mathrm{\infty }}{Lt}\frac{u_n}{v_n}=c$

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

P- series test: 

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

u_n = $\underset{n\to \mathrm{\infty }}{Lt}\sqrt{\frac{n}{n+1}}=1$ 

⇒ necessary condition fails 

Now, by limit comparison test 

Take 

$\underset{n\to \mathrm{\infty }}{Lt}\frac{u_n}{v_n}=\underset{n\to \mathrm{\infty }}{Lt}\sqrt{\frac{1}{1+\frac{1}{n}}}=1$ (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:

$log(1-x)=(-\mathrm{x}-\frac{{\mathrm{x}}^2}{2}-\frac{{\mathrm{x}}^3}{3}-\frac{{\mathrm{x}}^4}{4}...\mathrm{\infty })$

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

Calculation:

We know that,

${\log }_{\mathrm{e}}\left(\frac{1-\mathrm{x}}{1+\mathrm{x}}\right)={\log }_{\mathrm{e}}(1-\mathrm{x})-{\log }_{\mathrm{e}}(1+\mathrm{x})$

$=(-\mathrm{x}-\frac{{\mathrm{x}}^2}{2}-\frac{{\mathrm{x}}^3}{3}-\frac{{\mathrm{x}}^4}{4}...\mathrm{\infty })-(\mathrm{x}-\frac{{\mathrm{x}}^2}{2}+\frac{{\mathrm{x}}^3}{3}-\frac{{\mathrm{x}}^4}{4}...)$

$=-2\mathrm{x}-2\frac{{\mathrm{x}}^3}{3}-2\frac{{\mathrm{x}}^5}{5}....$

${\log }_{\mathrm{e}}\left(\frac{1-\mathrm{x}}{1+\mathrm{x}}\right)=-2[\mathrm{x}+\frac{{\mathrm{x}}^3}{3}+\frac{{\mathrm{x}}^5}{5}+...]$

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

Therefore,

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

$=\frac{-1}{2}{\log }_e\frac{1-\left(\frac{m-n}{m+n}\right)}{1+\left(\frac{m-n}{m+n}\right)}$

$=\frac{-1}{2}{\log }_{\mathrm{e}}\left(\frac{\mathrm{m}+\mathrm{n}-(\mathrm{m}-\mathrm{n})}{\mathrm{m}+\mathrm{n}+\mathrm{m}-\mathrm{n}}\right)$

$=\frac{-1}{2}{\log }_{\mathrm{e}}\left(\frac{2\mathrm{n}}{2\mathrm{m}}\right)=\frac{-1}{2}{\log }_{\mathrm{e}}\left(\frac{\mathrm{m}}{\mathrm{n}}\right)$

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

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 = $\underset{n\to \mathrm{\infty }}{Lt}\frac{u_{n+1}}{u_n}$

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 ;

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

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

∴ By ratio test ∑u_n, is convergent

Concept used:

Ratio test:

L = $\underset{n\to \mathrm{\infty }}{Lt}\frac{u_{n+1}}{u_n}$

P - Series test: 

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

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

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

∴ 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:

${\sum }_{n=1}^{\mathrm{\infty }}\sin \left(\frac{1}{n}\right)$

Concept used:

Limit Comparision test:

if an and bn are two positive series such that $\underset{n\to \mathrm{\infty }}{Lt}\frac{u_n}{v_n}=c$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:

${\sum }_{n=1}^{\mathrm{\infty }}\sin \left(\frac{1}{n}\right)$

take v_{n =} 

$\underset{n\to \mathrm{\infty }}{Lt}\frac{u_n}{v_n}=\underset{n\to \mathrm{\infty }}{Lt}\frac{\sin \left(\frac{1}{n}\right)}{\left(\frac{1}{n}\right)}=\underset{t\to 0}{Lt}\frac{\sin t}{t}$ (where t = 1/n) = 1

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

(p-series test, p = 1);

∴ ∑u_n is divergent.

Concept:

By ratio test, if $\underset{n\to \mathrm{\infty }}{lim}\frac{u_n}{u_{n+1}}=\lambda ,$

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

By Raabe’s test, if $\underset{n\to \mathrm{\infty }}{lim}n\left[\frac{u_n}{u_{n+1}}-1\right]=k,$

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,

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

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

When x² = 4,

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

$\underset{n\to \mathrm{\infty }}{lim}n\left[\frac{u_n}{u_{n+1}}-1\right]=-\frac{1}{2}<1$

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

Given:

${\mathrm{u}}_{\mathrm{n}}=\left(\frac{1}{\mathrm{n}}\right).\sin \left(\frac{1}{\mathrm{n}}\right)$

Concept used:

Limit Comparision test:

if an and bn are two positive series such that $\underset{n\to \mathrm{\infty }}{Lt}\frac{a_n}{b_n}=c$ 

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.

${\displaystyle \mathrm{L}{\mathrm{t}}_{\mathrm{n}\to \mathrm{\infty }}\left(\frac{{\mathrm{u}}_{\mathrm{n}}}{{\mathrm{v}}_{\mathrm{n}}}\right)={\displaystyle \mathrm{L}{\mathrm{t}}_{\mathrm{n}\to \mathrm{\infty }}\frac{\sin \left(\frac{1}{\mathrm{n}}\right)}{\left(\frac{1}{\mathrm{n}}\right)}}}$

$={\displaystyle \mathrm{L}{\mathrm{t}}_{\mathrm{n}\to 0}\left(\frac{\sin \mathrm{t}}{\mathrm{t}}\right)}$

where t = 1/n, Thus ${\displaystyle \mathrm{L}{\mathrm{t}}_{\mathrm{n}\to \mathrm{\infty }}\left(\frac{{\mathrm{u}}_{\mathrm{n}}}{{\mathrm{v}}_{\mathrm{n}}}\right)=1\ne 0}$

∴ By Limit comparison test Σu_n is convergent.