Convergence and Divergence of Series MCQ Quiz - Objective Question with Answer for Convergence and Divergence of Series - Download Free PDF

Concept:

Alternating series test:

Series of the form S = $\sum (-1{)}^n{a}_n$ or S = $\sum (-1{)}^{n+1}{a}_n$ is called alternating series. The series converges if the following conditions satisfy:

• ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{a}_n=0}$
• a_n+1 ≤ a_n 

Calculation:

Given series S = $\frac{1}{1^{\mathrm{p}}}-\frac{1}{2^{\mathrm{p}}}+\frac{1}{3^{\mathrm{p}}}-\frac{1}{4^{\mathrm{p}}}+...$

⇒ S = $\sum _{n=1}^{\mathrm{\infty }}(-1{)}^{n+1}\frac{1}{n^p}$

∴ a_n = $\frac{1}{n^p}$

Now, ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{a}_n}$ = ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\frac{1}{n^p}}$ = 0

Now, $\frac{a_{n+1}}{a_n}$

= $\frac{\frac{1}{(n+1{)}^p}}{\frac{1}{n^p}}$

= ${\left(\frac{n}{n+1}\right)}^p$

⇒ a_n+1 ≤ a_n, if p > 0

Hence, the series $\frac{1}{1^{\mathrm{p}}}-\frac{1}{2^{\mathrm{p}}}+\frac{1}{3^{\mathrm{p}}}-\frac{1}{4^{\mathrm{p}}}+...$ converges for p > 0.

concept- 

If series is $\sum a_n{x}^n$  then to find radius of convergence we have two formulas 

1. $\frac{1}{R}=li{m}_{n\to \mathrm{\infty }}\frac{a_{n+1}}{a_n}$ 

2. $\frac{1}{R}=li{m}_{n\to \mathrm{\infty }}(a_n{)}^{\frac{1}{n}}$

Explanation -

Here, Given series is $1+x+\frac{x^2}{2!}+.....$ 

we can write as $\sum _{n=0}^{\mathrm{\infty }}\frac{x^n}{n!}$ 

Here, $a_n=\frac{1}{n!}$  

Using Ratio Test,

 $\frac{1}{R}=li{m}_{n\to \mathrm{\infty }}\frac{a_{n+1}}{a_n}=\frac{(n!)}{(n+1)!}$  

$\frac{1}{R}=li{m}_{n\to \mathrm{\infty }}\frac{1}{(n+1)}=0$

Here, R = $\mathrm{\infty }$

Therefore, Correct Option is Option 1).

concept- 

If series is $\sum a_n{x}^n$  then to find radius of convergence we have two formulas 

1. $\frac{1}{R}=li{m}_{n\to \mathrm{\infty }}\frac{a_{n+1}}{a_n}$ 

2. $\frac{1}{R}=li{m}_{n\to \mathrm{\infty }}(a_n{)}^{\frac{1}{n}}$ 

Explanation- 

Here given series $1+x+x^2+x^3+.......$ 

and series be like $\sum _{n=0}^{\mathrm{\infty }}{x}^n$ 

Here, $a_n=1$ 

So,  $\frac{1}{R}=\underset{n\to \mathrm{\infty }}{lim}\frac{a_{n+1}}{a_n}$ = 1

Radius of Convergence is 1. 

Therefore, Correct Option is Option 1).

Concept:

Alternating series test:

Series of the form S = $\sum (-1{)}^n{a}_n$ or S = $\sum (-1{)}^{n+1}{a}_n$ is called alternating series. The series converges if the following conditions satisfy:

• ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{a}_n=0}$
• a_n+1 ≤ a_n 

Calculation:

Given series S = $\frac{1}{1^{\mathrm{p}}}-\frac{1}{2^{\mathrm{p}}}+\frac{1}{3^{\mathrm{p}}}-\frac{1}{4^{\mathrm{p}}}+...$

⇒ S = $\sum _{n=1}^{\mathrm{\infty }}(-1{)}^{n+1}\frac{1}{n^p}$

∴ a_n = $\frac{1}{n^p}$

Now, ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{a}_n}$ = ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\frac{1}{n^p}}$ = 0

Now, $\frac{a_{n+1}}{a_n}$

= $\frac{\frac{1}{(n+1{)}^p}}{\frac{1}{n^p}}$

= ${\left(\frac{n}{n+1}\right)}^p$

⇒ a_n+1 ≤ a_n, if p > 0

Hence, the series $\frac{1}{1^{\mathrm{p}}}-\frac{1}{2^{\mathrm{p}}}+\frac{1}{3^{\mathrm{p}}}-\frac{1}{4^{\mathrm{p}}}+...$ converges for p > 0.

Explanation -

Given that -

Let ΣUn be a convergent series of positive terms and let ΣVn be a divergent series of positive terms. 

so U_n is converges to 0 because ΣUn be a convergent series of positive terms.

So option(1) is true.

ΣVn be a divergent series of positive terms. 

let V_n = 1/n

Clearly, ΣVn is divergent series but V_n is convergent sequence and converges to 0.

Hence Option (2) and (3) are false.

let Vn = 3 + (-1)ⁿ

Then the ΣVn is divergent series but Vn is not convergent sequence.

Hence Option(4) is false.

Concept:

Alternating series test:

Series of the form S = $\sum (-1{)}^n{a}_n$ or S = $\sum (-1{)}^{n+1}{a}_n$ is called alternating series. The series converges if the following conditions satisfy:

• ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{a}_n=0}$
• a_n+1 ≤ a_n 

Calculation:

Given series S = $\frac{1}{1^{\mathrm{p}}}-\frac{1}{2^{\mathrm{p}}}+\frac{1}{3^{\mathrm{p}}}-\frac{1}{4^{\mathrm{p}}}+...$

⇒ S = $\sum _{n=1}^{\mathrm{\infty }}(-1{)}^{n+1}\frac{1}{n^p}$

∴ a_n = $\frac{1}{n^p}$

Now, ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{a}_n}$ = ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\frac{1}{n^p}}$ = 0

Now, $\frac{a_{n+1}}{a_n}$

= $\frac{\frac{1}{(n+1{)}^p}}{\frac{1}{n^p}}$

= ${\left(\frac{n}{n+1}\right)}^p$

⇒ a_n+1 ≤ a_n, if p > 0

Hence, the series $\frac{1}{1^{\mathrm{p}}}-\frac{1}{2^{\mathrm{p}}}+\frac{1}{3^{\mathrm{p}}}-\frac{1}{4^{\mathrm{p}}}+...$ converges for p > 0.

concept- 

If series is $\sum a_n{x}^n$  then to find radius of convergence we have two formulas 

1. $\frac{1}{R}=li{m}_{n\to \mathrm{\infty }}\frac{a_{n+1}}{a_n}$ 

2. $\frac{1}{R}=li{m}_{n\to \mathrm{\infty }}(a_n{)}^{\frac{1}{n}}$

Explanation -

Here, Given series is $1+x+\frac{x^2}{2!}+.....$ 

we can write as $\sum _{n=0}^{\mathrm{\infty }}\frac{x^n}{n!}$ 

Here, $a_n=\frac{1}{n!}$  

Using Ratio Test,

 $\frac{1}{R}=li{m}_{n\to \mathrm{\infty }}\frac{a_{n+1}}{a_n}=\frac{(n!)}{(n+1)!}$  

$\frac{1}{R}=li{m}_{n\to \mathrm{\infty }}\frac{1}{(n+1)}=0$

Here, R = $\mathrm{\infty }$

Therefore, Correct Option is Option 1).

concept- 

If series is $\sum a_n{x}^n$  then to find radius of convergence we have two formulas 

1. $\frac{1}{R}=li{m}_{n\to \mathrm{\infty }}\frac{a_{n+1}}{a_n}$ 

2. $\frac{1}{R}=li{m}_{n\to \mathrm{\infty }}(a_n{)}^{\frac{1}{n}}$ 

Explanation- 

Here given series $1+x+x^2+x^3+.......$ 

and series be like $\sum _{n=0}^{\mathrm{\infty }}{x}^n$ 

Here, $a_n=1$ 

So,  $\frac{1}{R}=\underset{n\to \mathrm{\infty }}{lim}\frac{a_{n+1}}{a_n}$ = 1

Radius of Convergence is 1. 

Therefore, Correct Option is Option 1).

Concept:

Alternating series test:

Series of the form S = $\sum (-1{)}^n{a}_n$ or S = $\sum (-1{)}^{n+1}{a}_n$ is called alternating series. The series converges if the following conditions satisfy:

• ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{a}_n=0}$
• a_n+1 ≤ a_n 

Calculation:

Given series S = $\frac{1}{1^{\mathrm{p}}}-\frac{1}{2^{\mathrm{p}}}+\frac{1}{3^{\mathrm{p}}}-\frac{1}{4^{\mathrm{p}}}+...$

⇒ S = $\sum _{n=1}^{\mathrm{\infty }}(-1{)}^{n+1}\frac{1}{n^p}$

∴ a_n = $\frac{1}{n^p}$

Now, ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{a}_n}$ = ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\frac{1}{n^p}}$ = 0

Now, $\frac{a_{n+1}}{a_n}$

= $\frac{\frac{1}{(n+1{)}^p}}{\frac{1}{n^p}}$

= ${\left(\frac{n}{n+1}\right)}^p$

⇒ a_n+1 ≤ a_n, if p > 0

Hence, the series $\frac{1}{1^{\mathrm{p}}}-\frac{1}{2^{\mathrm{p}}}+\frac{1}{3^{\mathrm{p}}}-\frac{1}{4^{\mathrm{p}}}+...$ converges for p > 0.

Concept:

Radius of convergent of a series  ${\displaystyle \sum _{\mathrm{n}=0}^{\infty }}$ anxn is given by $\frac{1}{R}=\underset{n\to \infty }{lim}\frac{a_{n+1}}{a_n}$

Explanation:

Here  an = (sin(n!)) / n! 

So $\frac{1}{R}=\underset{n\to \infty }{lim}\frac{a_{n+1}}{a_n}$

$\frac{1}{R}$ = $\underset{n\to \infty }{lim}\frac{sin(n+1)!}{(n+1)!}\times \frac{n!}{sinn!}$

$\frac{1}{R}=0$ ⇒ R = ∞ 

Hence options (1), (2) and (4) are correct.

Concept:

Alternating series test:

Series of the form S = $\sum (-1{)}^n{a}_n$ or S = $\sum (-1{)}^{n+1}{a}_n$ is called alternating series. The series converges if the following conditions satisfy:

• ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{a}_n=0}$
• a_n+1 ≤ a_n 

Calculation:

Given series S = $\frac{1}{1^{\mathrm{p}}}-\frac{1}{2^{\mathrm{p}}}+\frac{1}{3^{\mathrm{p}}}-\frac{1}{4^{\mathrm{p}}}+...$

⇒ S = $\sum _{n=1}^{\mathrm{\infty }}(-1{)}^{n+1}\frac{1}{n^p}$

∴ a_n = $\frac{1}{n^p}$

Now, ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{a}_n}$ = ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\frac{1}{n^p}}$ = 0

Now, $\frac{a_{n+1}}{a_n}$

= $\frac{\frac{1}{(n+1{)}^p}}{\frac{1}{n^p}}$

= ${\left(\frac{n}{n+1}\right)}^p$

⇒ a_n+1 ≤ a_n, if p > 0

Hence, the series $\frac{1}{1^{\mathrm{p}}}-\frac{1}{2^{\mathrm{p}}}+\frac{1}{3^{\mathrm{p}}}-\frac{1}{4^{\mathrm{p}}}+...$ converges for p > 0.

Explanation:

Here, it is given that

${\mathrm{u}}_{\mathrm{n}}=\sqrt{\mathrm{n}+1}-\sqrt{\mathrm{n}}$

and ${\mathrm{v}}_{\mathrm{n}}=\sqrt{{\mathrm{n}}^4+1}-\sqrt{{\mathrm{n}}^2}$

${\mathrm{u}}_{\mathrm{n}}=\sqrt{\mathrm{n}+1}-\sqrt{\mathrm{n}}$

$\frac{\sqrt{\mathrm{n}+1}-\sqrt{\mathrm{n}}}{\sqrt{\mathrm{n}+1}+\sqrt{\mathrm{n}}}\times \sqrt{\mathrm{n}+1}+\sqrt{\mathrm{n}}$

$=\frac{(\sqrt{\mathrm{n}+1}{)}^2-(\sqrt{\mathrm{n}}{)}^2}{\sqrt{\mathrm{n}+1}+\sqrt{\mathrm{n}}}=\frac{\mathrm{n}+1-\mathrm{n}}{\sqrt{\mathrm{n}+1}+\sqrt{\mathrm{n}}}=\frac{1}{\sqrt{\mathrm{n}+1}+\sqrt{\mathrm{n}}}$

Now, taking auxiliary series ${\mathrm{v}}_{\mathrm{n}}=\frac{1}{\sqrt{\mathrm{n}}}$

⇒ ${\displaystyle \underset{\mathrm{n}\to \mathrm{\infty }}{lim}\frac{{\mathrm{u}}_{\mathrm{n}}}{{\mathrm{v}}_{\mathrm{n}}}=\frac{\frac{1}{\sqrt{\mathrm{n}+1}+\sqrt{\mathrm{n}}}}{1/\sqrt{\mathrm{n}}}=\frac{\sqrt{\mathrm{n}}}{\sqrt{\mathrm{n}+1}+\sqrt{\mathrm{n}}}=1}$

finite and non-zero.

∴ $\Sigma {\mathrm{v}}_{\mathrm{n}}=\frac{1}{{\mathrm{n}}^{1/2}}$ is a p series as $\mathrm{p}=\frac{1}{2}$

Hence, Σv_n is divergent and Σu_n is also divergent and 

${\mathrm{v}}_{\mathrm{n}}=\sqrt{{\mathrm{n}}^4+1}-\sqrt{{\mathrm{n}}^2}$

$=\frac{\sqrt{{\mathrm{n}}^4+1}-{\mathrm{n}}^2}{\sqrt{{\mathrm{n}}^4+1}+{\mathrm{n}}^2}\times \sqrt{{\mathrm{n}}^4+1}+{\mathrm{n}}^2$

$=\frac{1}{\sqrt{{\mathrm{n}}^4+1}+{\mathrm{n}}^2}$

Now, taking auxiliary series $\Sigma {\mathrm{v}}_{\mathrm{n}}=\frac{1}{{\mathrm{n}}^2}$

∵ ${\displaystyle \underset{\mathrm{n}\to \mathrm{\infty }}{lim}\frac{{\mathrm{u}}_{\mathrm{n}}}{{\mathrm{v}}_{\mathrm{n}}}={\displaystyle \underset{\mathrm{n}\to \mathrm{\infty }}{lim}\frac{{\mathrm{n}}^2}{\sqrt{{\mathrm{n}}^4+1}+{\mathrm{n}}^2}=1}}$

finite and non-zero

∵ $\Sigma {\mathrm{v}}_{\mathrm{n}}=\frac{1}{{\mathrm{n}}^2}$ is a p-series at p = 2

Σv_n is convergent and 

Hence, Σu_n is also convergent.

Explanation:

Let the given series be denoted by Σu_n, then

Σ|u_n| = $\frac{2}{1^2}+\frac{3}{2^2}+\frac{4}{3^2}+\frac{5}{4^2}+....$

= $\Sigma \frac{\mathrm{n}+1}{{\mathrm{n}}^2}=\Sigma {\mathrm{v}}_{\mathrm{n}}$ (say)

Compare this series Σv_n with the auxiliary series,

Σw_n = Σ1/n

Then, ${\displaystyle \underset{\mathrm{n}\to \mathrm{\infty }}{lim}\frac{{\mathrm{v}}_{\mathrm{n}}}{{\mathrm{w}}_{\mathrm{n}}}={\displaystyle \underset{\mathrm{n}\to \mathrm{\infty }}{lim}[\frac{\mathrm{n}+1}{{\mathrm{n}}^2}\times \frac{\mathrm{n}}{1}]={\displaystyle \underset{\mathrm{n}\to \mathrm{\infty }}{lim}(1+\frac{1}{\mathrm{n}})=1}}}$

Which is finite quantity.

Hence Σu_n and Σv_n are either both convergent or both divergent.

But Σw_n = Σ$\frac{1}{\mathrm{n}}$ is divergent as the series Σ$\frac{1}{{\mathrm{n}}^2}$ divergent. if ρ = 1

Hence, the series Σv_n is divergent.

Also, in the series Σu_n, we find that its term are alternately positive and negative,

its terms are continuously decreasing and

${\displaystyle \underset{\mathrm{n}\to \mathrm{\infty }}{lim}{\mathrm{u}}_{\mathrm{n}}={\displaystyle \underset{\mathrm{n}\to \mathrm{\infty }}{lim}\left(\frac{\mathrm{n}+1}{{\mathrm{n}}^2}\right)}}$

$={\displaystyle \underset{\mathrm{n}\to \mathrm{\infty }}{lim}(\frac{1}{\mathrm{n}}+\frac{1}{{\mathrm{n}}^2})=0}$

Thus, all condition of Leibnitz's test are satisfied and as Σu_n is convergent.

Hence, the given series Σu_n is conditionally convergent.

Concept:

Convergence of series:

For any sequence {a_n} of positive terms, such that ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{a}_n}$ = 0, Σan is a convergent series .

Calculation:

Given that Σan is a convergent series of positive terms.

We know that the sequence {an} approaches zero as n approaches infinity. This is a necessary condition for the convergence of a series of positive terms.

Now, let's consider the series $\sum \frac{a_n}{1+a_n}$ ​ ​ 

= $\sum \frac{a_n\times \frac{1}{a_n}}{(1+a_n)\times \frac{1}{a_n}}$

= $\sum \frac{1}{1+\frac{1}{a_n}}$

Since, ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{a}_n}$ = 0, as  Σan​ converges, we have ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\frac{1}{1+\frac{1}{a_n}}}$ = 0.

Therefore, each term in the series $\sum \frac{a_n}{1+a_n}$ ​approaches zero as n approaches infinity.

∴ The series Σan / (1 + an)​ converges.

Correct answer is Option (3)

Explanation -

Given that -

Let ΣUn be a convergent series of positive terms and let ΣVn be a divergent series of positive terms. 

so U_n is converges to 0 because ΣUn be a convergent series of positive terms.

So option(1) is true.

ΣVn be a divergent series of positive terms. 

let V_n = 1/n

Clearly, ΣVn is divergent series but V_n is convergent sequence and converges to 0.

Hence Option (2) and (3) are false.

let Vn = 3 + (-1)ⁿ

Then the ΣVn is divergent series but Vn is not convergent sequence.

Hence Option(4) is false.