Infinite Series MCQ Quiz - Objective Question with Answer for Infinite Series - Download Free PDF

Calcu;ation: 

 ${\mathrm{Lim}}_{\mathrm{n}\to \mathrm{\infty }}\frac{0+3+6+9+\dots \dots \mathrm{n}\text{ terms }}{\sqrt{2{\mathrm{n}}^4+4\mathrm{n}+3}-\sqrt{{\mathrm{n}}^4+5\mathrm{n}+4}}$

⇒ ${\mathrm{Lim}}_{\mathrm{n}\to \mathrm{\infty }}\frac{3\mathrm{n}(\mathrm{n}-1)}{2(\sqrt{2{\mathrm{n}}^4+4\mathrm{n}+3}-\sqrt{{\mathrm{n}}^4+5\mathrm{n}+4})}$

= $\frac{3}{2(\sqrt{2}-1)}=\frac{3}{2}(\sqrt{2}+1)$

Hence, the correct answer is Option 3.

Calculation

Denominator :

$(n+1{)}^{a-1}[(na+1)+(na+2)+\dots +(na+n)]$

⇒ $(n+1{)}^{a-1}[n^{2}a+n(n+1)/2]$

⇒ $(n+1{)}^{a-1}[(2a+1)n^2+n]/2$

$\underset{n\to \mathrm{\infty }}{lim}(1^a+2^a+3^a+\dots +n^a)=\underset{n\to \mathrm{\infty }}{lim}{\int }_0^n{x}^{a}dx$

⇒ $\frac{1}{a+1}\underset{n\to \mathrm{\infty }}{lim}{n}^{a+1}$

$L=\underset{n\to \mathrm{\infty }}{lim}\frac{(1^a+2^a+\dots +n^a)}{(n+1{)}^{a-1}[(na+1)+(na+2)+\dots +(na+n)]}$

⇒ $\frac{2}{a+1}\underset{n\to \mathrm{\infty }}{lim}\frac{n^{a+1}}{(n+1{)}^{a-1}[(2a+1)n^2+n]}$

⇒ $L=\frac{2}{a+1}\underset{n\to \mathrm{\infty }}{lim}\frac{{\left(\frac{n}{n+1}\right)}^{a+1}}{(2a+1){\left(\frac{n}{n+1}\right)}^2+\frac{n}{(n+1{)}^2}}$

As $\underset{n\to \mathrm{\infty }}{lim}{\left(\frac{n}{n+1}\right)}^r=\underset{n\to \mathrm{\infty }}{lim}{\left(\frac{1}{1+\frac{1}{n}}\right)}^r=1$ and $\underset{n\to \mathrm{\infty }}{lim}\frac{n}{(n+1{)}^r}=0\text{if}r>1$

⇒ L = $\frac{2}{(a+1)(2a+1)}=\frac{1}{60}$

$\Rightarrow (a+1)(2a+1)=120$

By solving the above quadratic equation we get two values of $a$

⇒ $a=7$ or $\frac{-17}{2}$

For $a<0$ numerator diverges

$\Rightarrow a=7$

Concept Used:

Sum of first n square terms is given by ${\displaystyle \frac{n(n+1)(2n+1)}{6}}$

and $\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}=0$

Calculation:

$\stackrel{Lt}{n\to \mathrm{\infty }}\frac{1^2+2^2+3^2+.......+n^2}{n^3}$

which can be rewritten as using the concept as

$limn\to \mathrm{\infty }{\displaystyle \frac{n(n+1)(2n+1)}{6n^3}}$

$\Rightarrow limn\to \mathrm{\infty }{\displaystyle \frac{n^3(1+\frac{1}{n})(2+\frac{1}{n})}{6n^3}}$

$\Rightarrow limn\to \mathrm{\infty }{\displaystyle \frac{(1+\frac{1}{n})(2+\frac{1}{n})}{6}}$

Taking the limit in the above expression gives,

⇒ (91 + 0)× (2 + 0)/6 = 2/6 = 1/3

Hence, the required limit is 1/3.

Concept Used:

Sum of first n square terms is given by ${\displaystyle \frac{n(n+1)(2n+1)}{6}}$

and $\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}=0$

Calculation:

$\stackrel{Lt}{n\to \mathrm{\infty }}\frac{1^2+2^2+3^2+.......+n^2}{n^3}$

which can be rewritten as using the concept as

$limn\to \mathrm{\infty }{\displaystyle \frac{n(n+1)(2n+1)}{6n^3}}$

$limn\to \mathrm{\infty }{\displaystyle \frac{(2n^3+3n^2+n)}{6n^3}}$

$\Rightarrow limn\to \mathrm{\infty }{\displaystyle \frac{n^3(2+\frac{3}{n}+\frac{1}{n^2})}{6n^3}}$

$\Rightarrow limn\to \mathrm{\infty }{\displaystyle \frac{(2+\frac{3}{n}+\frac{1}{n^2})}{6}}$

Taking the limit in the above expression gives,

⇒ (2 + 0 + 0)/6 = 1/3

Hence, the required limit is 1/3.

Concept:

$ta{n}^{-1}\left(\frac{x-y}{1+xy}\right)=ta{n}^{-1}x-ta{n}^{-1}y$

Solution:

$S_k=\sum _{r=1}^k{\tan }^{-1}(\frac{6^r}{2^{2r+1}+3^{2r+1}})$

Divide by 3^2r

$S_k=\sum _{r=1}^{k}ta{n}^{-1}\left(\frac{{\left(\frac{2}{3}\right)}^r}{{\left(\frac{2}{3}\right)}^{2r}.2+3}\right)$

$S_k=\sum _{r=1}^{k}ta{n}^{-1}\left(\frac{{\left(\frac{2}{3}\right)}^r}{3({\left(\frac{2}{3}\right)}^{2r+1}+1)}\right)$

Let ${\left(\frac{2}{3}\right)}^r=t$

$S_k=\sum _{r=1}^{k}ta{n}^{-1}\left(\frac{\frac{t}{3}}{1+\frac{2}{3}{t}^2}\right)$

$S_k=\sum _{r=1}^{k}ta{n}^{-1}\left(\frac{t-\frac{2t}{3}}{1+t.\frac{2t}{3}}\right)$

Using Formula $ta{n}^{-1}\left(\frac{x-y}{1+xy}\right)=ta{n}^{-1}x-ta{n}^{-1}y$

$S_k=\sum _{r=1}^k(ta{n}^{-1}t-ta{n}^{-1}\left(\frac{2t}{3}\right))$

Put $t={\left(\frac{2}{3}\right)}^r$

$S_k=\sum _{r=1}^k(ta{n}^{-1}{\left(\frac{2}{3}\right)}^r-ta{n}^{-1}{\left(\frac{2}{3}\right)}^{r+1})$

$S_k=ta{n}^{-1}\left(\frac{2}{3}\right)-ta{n}^{-1}{\left(\frac{2}{3}\right)}^2+ta{n}^{-1}{\left(\frac{2}{3}\right)}^2-ta{n}^{-1}{\left(\frac{2}{3}\right)}^3+ta{n}^{-1}{\left(\frac{2}{3}\right)}^3-ta{n}^{-1}{\left(\frac{2}{3}\right)}^4+...+ta{n}^{-1}{\left(\frac{2}{3}\right)}^{k-1}-ta{n}^{-1}{\left(\frac{2}{3}\right)}^k+ta{n}^{-1}{\left(\frac{2}{3}\right)}^k-ta{n}^{-1}{\left(\frac{2}{3}\right)}^{k+1}$

$S_k=ta{n}^{-1}\left(\frac{2}{3}\right)-ta{n}^{-1}{\left(\frac{2}{3}\right)}^{k+1}$

${\displaystyle \underset{k\to \mathrm{\infty }}{lim}{S}_k={\displaystyle \underset{k\to \mathrm{\infty }}{lim}(ta{n}^{-1}\left(\frac{2}{3}\right)-ta{n}^{-1}{\left(\frac{2}{3}\right)}^{k+1})}}$

= $ta{n}^{-1}\left(\frac{2}{3}\right)-ta{n}^{-1}(0)$     $(\because 0<\frac{2}{3}<1,{\displaystyle \underset{k\to \mathrm{\infty }}{lim}{\left(\frac{2}{3}\right)}^{k+1}=0})$

∴ ${\displaystyle \underset{k\to \mathrm{\infty }}{lim}{S}_k=ta{n}^{-1}\left(\frac{2}{3}\right)=co{t}^{-1}\left(\frac{3}{2}\right)}$

So, The correct option is (2)

Concept Used:

Sum of first n square terms is given by ${\displaystyle \frac{n(n+1)(2n+1)}{6}}$

and $\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}=0$

Calculation:

$\stackrel{Lt}{n\to \mathrm{\infty }}\frac{1^2+2^2+3^2+.......+n^2}{n^3}$

which can be rewritten as using the concept as

$limn\to \mathrm{\infty }{\displaystyle \frac{n(n+1)(2n+1)}{6n^3}}$

$limn\to \mathrm{\infty }{\displaystyle \frac{(2n^3+3n^2+n)}{6n^3}}$

$\Rightarrow limn\to \mathrm{\infty }{\displaystyle \frac{n^3(2+\frac{3}{n}+\frac{1}{n^2})}{6n^3}}$

$\Rightarrow limn\to \mathrm{\infty }{\displaystyle \frac{(2+\frac{3}{n}+\frac{1}{n^2})}{6}}$

Taking the limit in the above expression gives,

⇒ (2 + 0 + 0)/6 = 1/3

Hence, the required limit is 1/3.

Concept:

$\underset{\mathrm{n}\to -\mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}\sum _{\mathrm{r}=1}^{\mathrm{p}\mathrm{n}}\frac{1}{\mathrm{n}}\mathrm{f}\left(\frac{\mathrm{r}}{\mathrm{n}}\right)={\int }_0^{\mathrm{p}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{d}\mathrm{x}$

Calculation:

The series given in the question is:

$\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}\left(\frac{{(\mathrm{n}+1)}^{1/3}}{{\mathrm{n}}^{4/3}}+\frac{{(\mathrm{n}+2)}^{1/3}}{{\mathrm{n}}^{4/3}}+\cdots \cdots +\frac{{(2\mathrm{n})}^{1/3}}{{\mathrm{n}}^{4/3}}\right)$

$=\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}\left(\frac{{(\mathrm{n}+1)}^{1/3}}{\mathrm{n}.{\mathrm{n}}^{1/3}}+\frac{{(\mathrm{n}+2)}^{1/3}}{\mathrm{n}.{\mathrm{n}}^{1/3}}+\cdots \cdots +\frac{{(2\mathrm{n})}^{1/3}}{\mathrm{n}.{\mathrm{n}}^{1/3}}\right)$

Taking n^(1/3) outside and cancelling the same in the denominator, we get

$=\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}\left(\frac{{\left(1+\frac{1}{\mathrm{n}}\right)}^{1/3}}{\mathrm{n}}+\frac{{\left(1+\frac{2}{\mathrm{n}}\right)}^{1/3}}{\mathrm{n}}+\cdots \cdots +\frac{{\left(1+\frac{\mathrm{n}}{\mathrm{n}}\right)}^{1/3}}{\mathrm{n}}\right)$

$=\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}\left({\left(1+\frac{1}{\mathrm{n}}\right)}^{1/3}+{\left(1+\frac{2}{\mathrm{n}}\right)}^{1/3}+\cdots \cdots +{\left(1+\frac{\mathrm{n}}{\mathrm{n}}\right)}^{1/3}\right)\frac{1}{\mathrm{n}}$

$=\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}\sum _{\mathrm{r}=1}^{\mathrm{n}}{\left(1+\frac{\mathrm{r}}{\mathrm{n}}\right)}^{1/3}\frac{1}{\mathrm{n}}$

Now taking, $\because \frac{\mathrm{r}}{\mathrm{n}}=\mathrm{x}\text{ and }\frac{1}{\mathrm{n}}=\text{dx}$

The equation becomes integral,

$=\underset{0}{\overset{1}{\int }}{\left(1+\mathrm{x}\right)}^{\frac{1}{3}}\mathrm{d}\mathrm{x}$

$={\left[\frac{{\left(1+\mathrm{x}\right)}^{\frac{4}{3}}}{\frac{4}{3}}\right]}_0^1$

$={\left[\frac{3}{4}{(1+\mathrm{x})}^{\frac{4}{3}}\right]}_0^1$

$=\frac{3}{4}{\left(1+1\right)}^{\frac{4}{3}}-\frac{3}{4}{\left(1+0\right)}^{\frac{4}{3}}$

$=\frac{3}{4}(2{)}^{\frac{4}{3}}-\frac{3}{4}$

Concept:

D' Alembert's test: If  Σu_n is a positive term series, such that,  $\underset{n\to \mathrm{\infty }}{lim}\frac{u_n}{u_{n+1}}=l$, then the series

(i) converges, if l < 1,

(ii) diverges, if l > 1,

(iii) the test fails, if l = 1

Additional Information

Comparison Test- If $\sum u_n$ and $\sum v_n$ are two positive term series, and k ≠ 0, a fixed positive real number and there exists a positive integer m such that u_n ≤ kv_n, 

∀ n ≥ m, then-

(i) $\sum u_n$ is convergent, if $\sum v_n$ is convergent and

(ii) $\sum v_n$ is divergent, if  $\sum u_n$  is divergent.

Raabe's Test-

If $\sum u_n$ is a positive term series, such that $\underset{n\to \mathrm{\infty }}{lim}n(\frac{u_n}{u_{n+1}}-1)=1$ then the series,

(i) converges, if l > 1,

(ii) diverges, if l < 1,

(iii) the test fails, if l = 1.

Given series is 1 + r + r2 + r3 + . . . . . ∞

1. |r| < 1 the series sum approaches to a fixed value hence it is convergent.
2. |r| > 1, the sum of series goes on increasing hence it is divergent.
3. r ≤ -1, the series does not approach a fixed value, hence it is oscillatory.

Concept:

Convergence and divergence of an Infinite series.

This is dependent on the convergence (or) divergence of the sequence of partial sums.

Let ${\displaystyle \sum _{k=1}^{\mathrm{\infty }}{U}_{k}bc}$ an in fenite series.

{S_n} be the sequence of partial sums.

Case (1)

If ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{S}_n=S}$, Where 'S' is a real number, then the infinite series is converges and ${\displaystyle \sum _{k=1}^{\mathrm{\infty }}{U}_K=S}$

Case (2)

If ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{S}_n}$ doesn't have a fenite limit, then series diverges.

If we can determine the whether limit of partial sum of an infinite series is fenite, then we know the infinite series converges else not.

Based on above definitions below table gives the common and recognizable infinite series.

Conclusion:

The Given series is p-series, So, it's divergent if p ≤ 1.

let p = -2

${\displaystyle \sum _{n=1}^{\mathrm{\infty }}{\displaystyle \frac{1}{n^{-2}}}=\sum _{n=1}^{\mathrm{\infty }}{n}^2=0+1+4+...\mathrm{\infty }}$   not finite value.

Concept:

Convergence and divergence of an Infinite series

This is dependent on the Convergence (or) divergence of the sequence of partial sums

Let $\sum _{K=1}^{\mathrm{\infty }}{U}_k$ be an infinite series.

{S_n} be the sequence of parial sums.

Case (1)

If $\underset{n\to \mathrm{\infty }}{lim}{S}_n=S$

where S is a real number, then the infinite series is converges and $\sum _{K=1}^{\mathrm{\infty }}{U}_k$

Case (II)

If $\underset{n\to \mathrm{\infty }}{lim}{S}_n$ doesn't have an finite limit, then series diverges.

If we can determine the whether limit of partial sum of an infinite series is finite, then we know the infinite series converges else not.

Based on the above definition below table gives the common and recognexable infinite series.

Calculation:

Given series is $1+\frac{2}{3}+{\left(\frac{2}{3}\right)}^2+...+{\left(\frac{2}{3}\right)}^{n-1}$ 

a = 1, r = 2/3

the given series is Geometric progression, From the table it's clear that common ratio |r| < 1 so the given series is convergent.

Alternative:

Summation value of this is

$S=\sum _{n=1}^{\mathrm{\infty }}a{r}^{n-1}$

$=\sum _{n=1}^{\mathrm{\infty }}1.{\left(\frac{2}{3}\right)}^{n-1}$

$=\frac{1}{1-\frac{2}{3}}$

$=\frac{1}{\frac{3-2}{3}}$

S = 3

It's finite value for infinite series.

So, it's convergent

Concept:

Series: If we add infinite terms of an infinite sequence ${\{a_n\}}_{n=1}^{\infty }$ , we get the expression in the form

⇒ a₁ + a₂ + a₃ + a₄ + ... + a_n + ...

• It is called an infinite series or series and is denoted for the shortly as $\sum _{n=1}^{\infty }{a}_n$ or $\sum a_n$

Convergent Series: If the given series {a_n} shows $li{m}_{n\to \infty }{a}_n=s$ and s exists as a real number and is finite and unique then this series $\sum a_n$ is called convergent series.

⇒ a1 + a2 + a3 + a4 + ... + an = s

∴ $\sum _{n=1}^{\infty }{a}_n$ = s

• simply in convergent series the sum of the series or the limit of the series, at last, give the real number or finite and unique then the series falls in the category of convergent series.
• Some examples of convergent series

$(\frac{1}{n})\to 0$

$(1+\frac{1}{n}{)}^n\to e$

$(\frac{1}{2}{)}^n\to 0$

​Divergent series: If the limit of the series {an} is infinite positive or infinite negative, the sequence is said to be divergent.

⇒ a1 + a2 + a3 + a4 + ... + an = $\pm \infty$

⇒ $li{m}_{n\to \infty }{a}_n=\pm \infty$

• some examples of divergent series

​-n³→ - ∞

3ⁿ→∞

(-1)^∞ → nowhere

(-2)^∞ → nowhere

Oscillatory series: If the limit of the series {a_n} is not unique, the sequence is said to be oscillatory.

$li{m}_{n\to \infty }{a}_n=notunique$

• some examples of oscillatory series

​{a_n} = $(-1{)}^n+\frac{1}{2^n}$ = 1 ,if n is even

                               = -1, if n is odd

∴ limit is not unique, the sequence is oscillatory.

• The simplest example of oscillatory series is (-1)ⁿ

^​

• If the value or limit of the series fluctuates and it continuously approaches the near to any fixed value then this series is called oscillatory convergent series.
• Example of oscillatory convergent series
• {a_n} = $\frac{(-1{)}^n}{n}$

Calculation:

Given Series:

⇒ $\frac{14}{1^3}+\frac{24}{2^3}+\frac{34}{3^3}+\frac{44}{4^3}+\frac{54}{5^3}+\frac{64}{6^3}+\dots$

⇒ $\frac{14}{1}+\frac{24}{8}+\frac{34}{27}+\frac{44}{64}+\frac{54}{125}+\frac{64}{216}...$

⇒ 14 + 3 + 1.25 + 0.6875 + 0.432 + 0.296 + ...

≈ 20

• As this series has a finite value of approximately 20, this series is convergent series.

Concept:

→ For an arithmetic-geometric series with the first term ‘a’, common ratio ‘r’, and common difference ‘d’, the sum to infinity is given by

$S_{\mathrm{\infty }}=\frac{a}{1-r}+\frac{dr}{{(1-r)}^2}$ ....(1)

Calculation:

Given:

Here, P (X = t) = 1

$\Rightarrow$ $\sum _{t=0}^{\mathrm{\infty }}\frac{(t+1)a}{3^t}$ = 1

$\Rightarrow$ a + $\frac{2a}{3}$ + $\frac{3a}{3^2}$ + --- + $\mathrm{\infty }$ = 1

$\Rightarrow$ a(1 + $\frac{2}{3}$ + $\frac{3}{3^2}$ + --- + $\mathrm{\infty }$) = 1

The term in the bracket is arithmetic-geometric series with first term 1, common ratio 1/3, and common difference 1.

From equation (1), the series can be rewritten as,

$\Rightarrow$ $\frac{1}{1-\frac{1}{3}}+\frac{1\times \frac{1}{3}}{{(1-\frac{1}{3})}^2}$  

$\Rightarrow$ $a(\frac{3}{2}+\frac{1}{3}\times \frac{9}{4})$ = 1

$\Rightarrow$ a($\frac{3}{2}$ + $\frac{1}{3}\times \frac{9}{4}$) = 1

$\Rightarrow$ $\frac{9a}{4}$ = 1

$\Rightarrow$ $\frac{1}{a}$ = $\frac{9}{4}$

Hence, the correct answer is option 4.

Concept Used:

Sum of first n square terms is given by ${\displaystyle \frac{n(n+1)(2n+1)}{6}}$

and $\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}=0$

Calculation:

$\stackrel{Lt}{n\to \mathrm{\infty }}\frac{1^2+2^2+3^2+.......+n^2}{n^3}$

which can be rewritten as using the concept as

$limn\to \mathrm{\infty }{\displaystyle \frac{n(n+1)(2n+1)}{6n^3}}$

$limn\to \mathrm{\infty }{\displaystyle \frac{(2n^3+3n^2+n)}{6n^3}}$

$\Rightarrow limn\to \mathrm{\infty }{\displaystyle \frac{n^3(2+\frac{3}{n}+\frac{1}{n^2})}{6n^3}}$

$\Rightarrow limn\to \mathrm{\infty }{\displaystyle \frac{(2+\frac{3}{n}+\frac{1}{n^2})}{6}}$

Taking the limit in the above expression gives,

⇒ (2 + 0 + 0)/6 = 1/3

Hence, the required limit is 1/3.

Concept:

$\underset{\mathrm{n}\to -\mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}\sum _{\mathrm{r}=1}^{\mathrm{p}\mathrm{n}}\frac{1}{\mathrm{n}}\mathrm{f}\left(\frac{\mathrm{r}}{\mathrm{n}}\right)={\int }_0^{\mathrm{p}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{d}\mathrm{x}$

Calculation:

The series given in the question is:

$\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}\left(\frac{{(\mathrm{n}+1)}^{1/3}}{{\mathrm{n}}^{4/3}}+\frac{{(\mathrm{n}+2)}^{1/3}}{{\mathrm{n}}^{4/3}}+\cdots \cdots +\frac{{(2\mathrm{n})}^{1/3}}{{\mathrm{n}}^{4/3}}\right)$

$=\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}\left(\frac{{(\mathrm{n}+1)}^{1/3}}{\mathrm{n}.{\mathrm{n}}^{1/3}}+\frac{{(\mathrm{n}+2)}^{1/3}}{\mathrm{n}.{\mathrm{n}}^{1/3}}+\cdots \cdots +\frac{{(2\mathrm{n})}^{1/3}}{\mathrm{n}.{\mathrm{n}}^{1/3}}\right)$