Informative line

### Different Tests On Series And P Series

Learn rules of convergence & divergence test series, integral test for convergence. Practice convergence and divergence of 'P' series.

# Theorem on Convergence of a Series, Test of Divergence

• If the series $$\sum\limits_{n=1} ^\infty\; a_n$$ converges then $$\lim\limits _{n\to \infty}\; a_n = 0$$
• This means that the terms of a converging series must be very close to 0 for very large values of n.
• This will not mean that if $$\lim\limits _{n\to \infty }\; a_n = 0$$ then series converges. If $$\lim\limits _{n\to \infty }\; a_n = 0$$ then the series may or may not converge.
• If $$\lim\limits _{n\to \infty }\; a_n$$  does not exist or if $$\lim\limits _{n\to \infty }\; a_n \neq 0$$  then series $$\sum\limits_{n=1}^\infty \; a_n$$ is divergent.

e.g.   Consider the series

$$\sum \limits _{n=1}^\infty\dfrac{n-2}{4n + 5} ,$$ this series is divergent as

$$\lim\limits _{n\to \infty} \; a_n = \lim\limits _{n\to \infty }\; \dfrac{n-2}{4n +5} = \lim\limits _{n\to \infty}\; \dfrac{1-\dfrac{2}{n}}{4+\dfrac{5}{n }}\;\;=\dfrac{1}{4}\; \neq\; 0$$

So if $$\lim\limits _{n\to \infty}\; a_n \neq 0$$ we say that the series $$\sum\limits ^\infty_{n=1}\; a_n$$ is divergent where as if $$\lim\limits _{n\to \infty }\; a_n = 0$$  nothing can be said about convergence or divergence of the series.

#### Which of the following series is divergent ?

A $$\sum\limits_{n=1}^\infty \;\; \left(\dfrac{1}{2}\right)^n$$

B $$\sum\limits_{n=1}^\infty \;\; 2 × \dfrac{3^n}{4^{n-1}}$$

C $$\sum\limits_{n=1}^\infty \;\; ln \left(\dfrac{3n^2+1}{4n^2+3}\right)$$

D $$\sum\limits_{n=1}^\infty \;\; \left(\dfrac{5}{7}\right)^n$$

×

If  $$\lim\limits _{n\to\infty}\; a_n \neq 0$$  then series $$\sum\limits _1^\infty\;a_n$$ will be divergent and geometric series is convergent if  |r| < 1

Consider the option

$$(a)\;\;\;\;\;\sum\limits _{n=1}^\infty\; \left(\dfrac{1}{2}\right)^n\; = \dfrac{1}{2} +\dfrac{1}{2^2} +\dfrac{1}{2^3}+ .....$$

This is a geometric series with  $$a= \dfrac{1}{2}, \; r= \dfrac{1}{2}$$

$$\therefore$$  it is convergent

$$(b)\;\;\;\;\sum \limits _{n=1}^\infty\; 2 × \dfrac{3^n}{4 ^ {n-1}} \;\; = \sum \limits ^\infty_{n=1}\; 8 × \dfrac{3^n}{4^n}\; = 8 \; \sum \limits _{n=1}^\infty\; \left(\dfrac{3}{4}\right)^n$$

This again is a geometric series with $$a = \dfrac{3}{4}$$

$$\therefore$$ It is convergent

$$(c)\;\;\;\;\; \sum\limits^\infty_{n=1}\; ln \left(\dfrac{3n^2 +1}{4n^2+3}\right)$$

$$\lim\limits _{n\to \infty}\; a_n = \lim\limits _{n\to \infty}\; ln \left(\dfrac{3n^2 +1}{4n^2 +3}\right) = \lim\limits _{n\to \infty}\; ln \dfrac{3+\dfrac{1}{n^2}}{4+\dfrac{3}{n^2}}$$

$$= ln \dfrac{3}{4} \neq 0$$

$$\therefore$$  Series is divergent

$$(d)\;\;\;\;\; \sum\limits_{n=1}^\infty\; \left(\dfrac{5}{7}\right)^n = \dfrac{5}{7}+ \left(\dfrac{5}{7}\right)^2+ \left(\dfrac{5}{7}\right)^3+ .....$$

This is a geometric series with $$a= \dfrac{5}{7}, \;\;r=\dfrac{5}{7}$$

$$\therefore$$  It is convergent

### Which of the following series is divergent ?

A

$$\sum\limits_{n=1}^\infty \;\; \left(\dfrac{1}{2}\right)^n$$

.

B

$$\sum\limits_{n=1}^\infty \;\; 2 × \dfrac{3^n}{4^{n-1}}$$

C

$$\sum\limits_{n=1}^\infty \;\; ln \left(\dfrac{3n^2+1}{4n^2+3}\right)$$

D

$$\sum\limits_{n=1}^\infty \;\; \left(\dfrac{5}{7}\right)^n$$

Option C is Correct

# Rules of Convergence

If $$\sum\ a_n$$ and $$\sum \; b_n$$ are convergent series then the following series are also convergent

$$(1)\;\;\;\; \sum \; ca_n$$ where c is a constant

$$(2)\;\;\;\; \sum \;(a_n +b_n)$$

$$(3)\;\;\;\; \sum \;(a_n-b_n)$$  and we have

$$\sum \limits ^\infty_{n=1}\;ca_n = c\sum\limits^\infty_{n=1}\; a_n$$

$$\sum \limits ^\infty_{n=1} (a_n +b_n) = \sum \limits ^\infty_{n=1}\; a_n + \sum \limits ^\infty_{n=1} b_n$$

$$\sum \limits ^\infty_{n=1} (a_n -b_n) = \sum \limits ^\infty_{n=1}\; a_n - \sum \limits ^\infty_{n=1} b_n$$

The above properties of convergent series follow directly for corresponding limits laws, for sequences.

#### Find the sum of the series $$\sum\limits^\infty_{n=2} \left(\dfrac{5}{n^2-1}+ \dfrac{1}{2^n}\right)$$

A $$\dfrac{5}{2}$$

B $$\dfrac{17}{4}$$

C $$\dfrac{13}{4}$$

D $$\dfrac{17}{6}$$

×

$$\sum\limits_{n=1}^\infty\; (a_n +b_n) = \sum\limits_{n=1}^\infty\; a_n + \sum\limits_{n=1}^\infty b_n$$

If $$\sum\;a_n$$ and $$\sum\;b_n$$ are both convergent

Consider the given series

$$\sum\limits_{n = 2}^\infty \left(\dfrac{5}{n^2 -1} +\dfrac{1}{2^n}\right)= \sum\limits_{n = 2}^\infty \dfrac {5}{n^2 -1 } +\sum\limits_{n = 2}^\infty \dfrac{1}{2^n}$$

Consider the first series

$$\sum\limits^\infty_{n = 2} \; \dfrac{5}{n^2-1}\;\;= 5 \sum\limits^\infty_{n=2} \; \dfrac{1}{n^2 -1}$$

$$a_n = \dfrac{1}{n^2-1} = \dfrac{1}{2} \left[\dfrac{1}{n-1}-\dfrac{1}{n+1}\right]$$

$$\therefore \;\; \sum\limits _{n=2}^\infty\; \dfrac{1}{n^2-1}$$

$$=\dfrac{1}{2} \left[\left(\dfrac{1}{1}-\dfrac{1}{3}\right) +\left(\dfrac{1}{2}-\dfrac{1}{4}\right) +\left(\dfrac{1}{3}-\dfrac{1}{5}\right)+...+...\left(\dfrac{1}{n-1}-\dfrac{1}{n+1}\right)+...\right]$$

$$\dfrac{1}{2} \underbrace {\left[1 + \dfrac{1}{2} - \dfrac{1}{n} - \dfrac{1}{n+1}\right]}_\text{All other terms cancel out}\; _\text{where}\;\;n\to \infty$$

$$= \dfrac{1}{2} \left(\dfrac{3}{2}\right)= \dfrac{3}{4}$$

$$\therefore\; \sum \limits ^\infty_{n=2} \;\;\dfrac{5}{n^2-1} = 5 × \dfrac{3}{4} = \dfrac{15}{4}$$

Consider the second series

$$\sum\limits _{n = 2}^\infty\; \dfrac{1}{2^n} = \dfrac{1}{2^2} + \dfrac{1}{2^3} + \dfrac{1}{2^4} + .....$$

This is a geometric series with $$a = \dfrac{1}{4}\; , r =\dfrac{1}{2}$$

$$\therefore$$ $$S = \dfrac{a}{1-r} = \dfrac{\dfrac{1}{4}}{1-\dfrac{1}{2}}\; = \dfrac{1}{2}$$

$$\therefore\; \sum\limits ^\infty_{n=2} \;\; \dfrac{5}{n^2 - 1} + \dfrac{1}{2^n}\; = \dfrac{15}{4} \; \; + \;\;\dfrac{1}{2}\;\;=\;\;\dfrac{17}{4}$$

### Find the sum of the series $$\sum\limits^\infty_{n=2} \left(\dfrac{5}{n^2-1}+ \dfrac{1}{2^n}\right)$$

A

$$\dfrac{5}{2}$$

.

B

$$\dfrac{17}{4}$$

C

$$\dfrac{13}{4}$$

D

$$\dfrac{17}{6}$$

Option B is Correct

# Estimating the Sum of Series

• After confirming that a series $$\sum\; a_n$$ is convergent by integral test, we are interested in finding an approximation to the sum s of of the series $$\sum\; a_n$$.
• Any partial sum Sn is an approximation to S, but how good is this approximation , to find this we find the remainder.

$$R_n =S-S_n =a_{n+2}+ a_{n+3}+......$$

Rn = error made when Sn is used as an approximate to the sum S.

• Consider the following figure  $$R_n = a_{n+1} + a_{n+2} + a_{n+3} + ...... \leq\; \int \limits_n^\infty f (x)\; dx$$

(Note that integral is shaded area + some extra area )

Also $$R_n = a_{n+1} + a_{n+2}+ a_{n+3}.....\geq \; \int\limits_{n+1}^\infty\; f (x)\;dx$$  $$\therefore\;\;\; \int\limits_{n+1}^\infty\; f(x)\; dx \; \leq \; R_n \leq \int\limits_n^\infty\; f (x) \; dx$$

• Suppose $$f(k) = a_k$$ where f is a continuous, positive decreasing function for $$x\geq n$$ and $$\sum\;a_n$$ is convergent if  $$R_n = S-S_n$$ then

$$\int\limits_{n+1}^\infty\; f(x) dx \; \leq\; R_n \leq \; \int \limits_n^\infty f (x) \; dx$$

#### For the series $$\sum\limits _{n=1}^\infty\; \dfrac{1}{n^2}$$ find a value of n such that Sn ( the partial sum to n terms ) is such that Rn = S–Sn is less than 0.001 ?

A n > 150

B n > 420

C n > 575

D n > 1000

×

If $$R_n = S -S_n$$ then

$$\int\limits_{n+1}^\infty\; f(x) dx \leq R_n \leq \; \int\limits_n^\infty \; f (x) \; dx$$

In this case $$f (x) = \dfrac{1}{x^2}$$

$$\therefore\;\; R_n \leq\; \int\limits^\infty_n \; \dfrac{1}{x^2}\; dx = \dfrac{-1}{x}\Big]^\infty_n\; =- \left[0- \dfrac{1}{n}\right] = \dfrac{1}{n}$$

Now we desire

$$R_n < 0.001 \;\;$$

$$\Rightarrow\;\; \dfrac{1}{n} < 0.001$$

$$\Rightarrow\;\; n > \dfrac{1}{0.001}$$

$$\Rightarrow\;\; n > 1000$$

### For the series $$\sum\limits _{n=1}^\infty\; \dfrac{1}{n^2}$$ find a value of n such that Sn ( the partial sum to n terms ) is such that Rn = S–Sn is less than 0.001 ?

A

n > 150

.

B

n > 420

C

n > 575

D

n > 1000

Option D is Correct

# Comparison Tests

• In comparison test we compare a given series with a series that is known to be convergent or divergent.
• Suppose $$\sum\;a_n$$ and $$\sum\; b_n$$ are series with positive terms.
• If $$\sum\; b_n$$ is convergent and $$a_n\leq b_n$$ for all n, then $$\sum\;a_n$$ is also convergent.
• If $$\sum\; b_n$$ is divergent and $$a_n\geq b_n$$ for all n, then $$\sum\;a_n$$ is also divergent.
• The above statement is called comparison test

e.g.   $$\sum\limits_{n=1}^\infty\; \dfrac{1}{3^n +1}$$ is convergent because $$\sum\limits _{n=1}^\infty \; \dfrac{1}{3^n}\;\;=\dfrac{1}{2}$$ is convergent and $$\dfrac{1}{3^n+1}\; <\; \dfrac{1}{3^n } \forall\; n.$$

#### Using comparison test, what can be said about the series  $$\sum\limits_{n=1}^\infty \; \left(\dfrac{8^n}{2+9^n}\right)$$ .

A The series is convergent

B The series is divergent

C $$S= \sum \limits_{n=1}^\infty\; \dfrac{8^n}{2+9^n}\; > 8$$

D The series is a geometric series

×

Suppose $$\sum\;a_n$$ and $$\sum\; b_n$$ are series with positive terms.

If $$\sum\; b_n$$ is convergent and $$a_n\leq b_n$$ for all n, then $$\sum\;a_n$$ is also convergent.

If $$\sum\; b_n$$ is divergent and $$a_n\geq b_n$$ for all n, then $$\sum\;a_n$$ is also divergent.

In this case $$a_n = \dfrac{8^n}{2 +9 ^n}$$, consider for all n.

$$b_n = \dfrac{8^n}{9^n} = \left(\dfrac{8}{9}\right)^n$$

clearly bn > an for all n.

$$\sum \limits_{n =1}^\infty \left(\dfrac{8}{9}\right)^n = \dfrac{8}{9} +\left(\dfrac{8}{9}\right)^2+.......$$

This is a geometric series with  $$S= \dfrac{8/9}{1-8/9} = 8$$

$$\left(r = 8/9, \;\;a=8/9,\;\;S= a/1-r\right)$$

$$\therefore\;\; \sum \; b_n$$ is convergent

By comparison test $$\sum\; a_n$$ is also convergent and $$S= \sum \limits _{n=1}^\infty\; a_n< 8$$

### Using comparison test, what can be said about the series  $$\sum\limits_{n=1}^\infty \; \left(\dfrac{8^n}{2+9^n}\right)$$ .

A

The series is convergent

.

B

The series is divergent

C

$$S= \sum \limits_{n=1}^\infty\; \dfrac{8^n}{2+9^n}\; > 8$$

D

The series is a geometric series

Option A is Correct

# The Integral Test for Convergence of Series

• In general the partial sum of most of series is very difficult to find so, if we do not have an expression of Sn we can not test for convergence of series.
• Integral test is a test that enables us to determine whether series is convergent or divergent without explicitly founding the sum.
• Consider the series

$$S= \dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+......$$

There is no formula for Sn . Look at Sn for very various values of n.

 n $$S_n = \sum \limits_{i = 1}^n\; 1/i^2$$ 5 1.4636 10 1.5498 100 1.6350 500 1.6429 1000 1.6439 5000 1.6447

If looks as if the series is convergent and converges to a value near 1.64.

Now consider the curve $$y =\dfrac{1}{x^2}$$  The series $$\dfrac{1}{1^2}+\dfrac{1}{2^2} +......=$$ sum of all shaded areas.

Now consider  $$I= \int\limits_1^\infty \; \dfrac{1}{x^2}\; dx = \dfrac{-1}{x}\Big]^\infty_1 = 1$$

This is area under the curve $$y = \dfrac{1}{x^2}$$  from 1 to $$\infty$$

$$\therefore$$ Series value $$<1+ \int\limits_1^\infty\;\dfrac{1}{x^2}\; dx$$

$$<1+1<2$$

$$\therefore$$ Series is convergent

This leads to the integral test

• Suppose f is a continuous, positive, decreasing function on $$[1,\infty]$$ and let $$a_n = f (x)$$. Then the series $$\sum\limits_{n=1}^\infty\; a_n$$ is convergent if and only if $$\int\limits_1^\infty\; f (x) \; dx$$  is convergent.

If $$\int\limits_1^\infty\; f (x) \; dx$$ is convergent then $$\sum\limits _{n= 1} ^ \infty\; a_n$$ is convergent.

If  $$\int\limits_1^\infty\; f (x) \; dx$$ is divergent then  $$\sum\limits _{n= 1} ^ \infty\; a_n$$ is divergent.

#### Using Integral  test, find which of the following series is convergent.

A $$\sum\limits_{n=1}^\infty\; \dfrac{n^2}{n^3 +1}$$

B $$\sum\limits_{n=1}^\infty\; \dfrac{1}{\sqrt{n+4}}$$

C $$\sum\limits_{n=1}^\infty\; n^2e^{-n^3}$$

D $$\sum\limits_{n=1}^\infty\; \dfrac{n}{n^2+1}$$

×

If $$\int\limits_1^\infty\; f (x) \; dx$$ is convergent then $$\sum\limits _{n= 1} ^ \infty\; a_n$$ is convergent.

If  $$\int\limits_1^\infty\; f (x) \; dx$$ is divergent then  $$\sum\limits _{n= 1} ^ \infty\; a_n$$ is divergent.

Consider the options

$$(a)\;\;\;\sum\limits_{n=1}^\infty\;\dfrac{n^2}{n^3+1}\; \rightarrow\; a_n = \dfrac{n^2}{n^3+1}\; \Rightarrow\; f (x) = \dfrac{x^2}{x^3+1}$$

$$I = \int\limits_1^\infty\; \dfrac{x^2}{x^3 +1}\; dx \; \rightarrow\; put \;x^ 3 +1=t$$

$$\Rightarrow \; 3x^2 \; dx\;=\;dt$$

when $$x=\infty\rightarrow\; t = \infty\;\;\\\;x=1 \rightarrow\; t=2$$

$$\therefore\; I = \dfrac{1}{3} \int \limits_2^\infty \; \dfrac{1}{t} \; dt= \dfrac{1}{3} \; lnt \Big ]^\infty_2\; =\infty$$

$$\therefore \; I$$ is divergent

$$\therefore$$ Series is also divergent

$$(b)\;\;\;\sum\limits_{n=1}^\infty \dfrac{1}{\sqrt{n+4}}\; \rightarrow a_n= \dfrac{1}{\sqrt{n+4}}$$

$$\Rightarrow\; f (x) = \dfrac{1}{\sqrt {x+4}}$$

$$I = \int\limits_1^\infty \; \dfrac{1}{\sqrt{x+4}} \; dx = \dfrac{(x+4)^{1/2}}{1/2}\Bigg ]_1^\infty = \infty$$

$$\therefore\; I$$ is divergent

$$\Rightarrow$$ Series is divergent

$$(c)\;\;\;\;\sum\limits_{n=1}^\infty \; n^2 e ^{-n^3}\;\Rightarrow \; a_n = n^2 e ^{-n^3}$$

$$\Rightarrow f (x) = x^2 e ^{-x^3}$$

$$I = \int\limits_1^\infty x^2 e ^{-x^3}\; dx \; \rightarrow \; put\; x^3 =t$$

$$\Rightarrow 3x^2 \; dx = dt$$

Where $$x = \infty, \; t=\infty\;\;\; x=1,\; t= 1$$

$$\therefore\;\; I = \dfrac{1}{3}\; \int \limits _1^\infty\; e^{-t} \; dt = \dfrac{1}{3} \; (-e^{-t})\Big]^\infty_1$$

$$= \dfrac{-1}{3} [0-e^{-1}] = \dfrac{1}{3e}$$

$$\therefore\; I$$ is convergent

$$\Rightarrow$$ Series is convergent

$$(d)\;\;\;\; \sum\limits_{n=1} ^\infty \;\; \dfrac{n}{n^2 + 1}\;\; \rightarrow\; a_n = \dfrac{n}{n^2 +1}= f(n)$$

$$\Rightarrow\; f(x) = \dfrac{x}{x^2 +1}$$

$$I = \int\limits_1^\infty\; \dfrac{x}{x^2 +1}\; dx\; \rightarrow \; put\; x^2 +1 = t$$

$$\Rightarrow\; 2x \; dx = dt$$

Where $$x = \infty, \;\; t = \infty\;,\;\;\;x=1, \; t=2$$

$$\therefore\;\;I = \dfrac{1}{2}\; \int \limits _2^\infty\; \dfrac{dt}{t} = \dfrac{1}{2}\; lnt \; \Big ]^\infty_2\;=\infty$$

$$\therefore\;\; I$$  is divergent

$$\Rightarrow$$ Series is divergent

$$\therefore$$ Correct option is 'C'

### Using Integral  test, find which of the following series is convergent.

A

$$\sum\limits_{n=1}^\infty\; \dfrac{n^2}{n^3 +1}$$

.

B

$$\sum\limits_{n=1}^\infty\; \dfrac{1}{\sqrt{n+4}}$$

C

$$\sum\limits_{n=1}^\infty\; n^2e^{-n^3}$$

D

$$\sum\limits_{n=1}^\infty\; \dfrac{n}{n^2+1}$$

Option C is Correct

# Convergence and Divergence of 'P' Series

• The series $$\sum\limits_{n=1}^\infty\; \dfrac{1}{n^p}$$ is called the p series.
• If p < 0 then $$\lim\limits_{n\to\infty}\; \dfrac{1}{n^p} = \infty$$, so the p series diverges by test of divergence.
• If p = 0 then $$\lim\limits_{n\to \infty}\; \dfrac{1}{n^p} 1 \neq 0$$, so the p series diverge by test of divergence.
• If p > 0 then $$f(x) = \dfrac{1}{x^p}$$ is continuous, positive and decreasing on $$[1, \infty]$$

$$\int\limits_1^\infty\; \dfrac{1}{x^p} = \dfrac{x^{-p+1}}{-p+1}\Bigg]_1^\infty\; =\infty\;\; if \; 0 <p<1$$

$$\therefore$$ By integral test p series converges for p > 1 and  diverges if $$p\leq 1$$

The p series $$\sum\limits_{n=1}^\infty\; \dfrac{1}{n^p}$$ is convergent if p > 1 and divergent if $$p\;\leq\;1$$.

#### Euler found the sum of p series with p = 4 $$\sum\limits_{n= 1}^\infty\; \dfrac{1}{n^4}\; =\dfrac{\pi^4}{90}$$ Assuming this result, find the value of  $$\sum\limits _{n=1}^\infty \left(\dfrac{3}{n}\right)^4$$

A $$\dfrac{8 \pi ^4}{5}$$

B $$\dfrac{9\pi ^4}{10}$$

C $$\dfrac{10\pi ^4}{9}$$

D $$\dfrac{7\pi ^4}{8}$$

×

The p series $$\sum\limits_{n=1}^\infty\; \dfrac{1}{n^p}$$ is convergent if p > 1 and divergent if $$p\;\leq\;1$$.

Given that $$\sum\limits_{n=1}^\infty\; \dfrac{1}{n^4} = \dfrac{\pi^4}{90}$$

$$\therefore\;\;\sum\limits_{n=1}^\infty\; \left(\dfrac{3}{n}\right)^4 = 81 \; \sum\limits_{n=1}^\infty\; \dfrac{1}{n^4}\;=81 × \dfrac{\pi^4}{90}\;=\; \dfrac{9 \pi^4}{10}$$

### Euler found the sum of p series with p = 4 $$\sum\limits_{n= 1}^\infty\; \dfrac{1}{n^4}\; =\dfrac{\pi^4}{90}$$ Assuming this result, find the value of  $$\sum\limits _{n=1}^\infty \left(\dfrac{3}{n}\right)^4$$

A

$$\dfrac{8 \pi ^4}{5}$$

.

B

$$\dfrac{9\pi ^4}{10}$$

C

$$\dfrac{10\pi ^4}{9}$$

D

$$\dfrac{7\pi ^4}{8}$$

Option B is Correct