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Alternating Series Ratio Root Test And Convergence Interval

Learn an alternating series as series whose terms are alternately positive or negative. Practice to find out which of the following series convergence & divergence using root test.

Alternating Series

We define an alternating series as series whose terms are alternately positive or negative.

e.g.      $$1-\dfrac{1}{2} +\dfrac{1}{3}-\dfrac{1}{4} + ......(-1)^{n-1}\dfrac{1}{n}....$$

or      $$\dfrac{1}{2} -\dfrac{1}{4}+\dfrac{1}{8} -......(-1)^{n-1}\dfrac{1}{2^n}....$$

are both alternating series.

• Observe that  $$n^{th}$$ term of an alternating series is of the form

$$a_n = (-1)^n \,b_n \,\,\,{\text{or }}\,\, a_n = (-1)^{n-1} \,b_n$$

where, $$b_n$$ is a  positive number.

• If the alternating series ,

$$\sum\limits^{n-1}_1 (-1)^{n-1} \,b_n = b_1-b_2+b_3-b_4.......b_n$$

satisfies

(a)   $$b_{n+1} \leq b_n$$  for all n

(b)   $$\lim\limits_{n\to\infty} b_n =0$$

then the series is convergent.

• The above result is called Alternating Series Test.

Using Alternating Series Test and test for divergence, choose the divergent series from the following

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

B $$\sum\limits_{n=1}^\infty (-1)^{n-1}\,n\,e^{-n}$$

C $$\sum\limits_{n=1}^\infty (-1)^{n}\,\dfrac{3n-1}{2n+1}$$

D $$\sum\limits_{n=1}^\infty (-1)^{n}\,\dfrac{sin\,\pi}{n}$$

×

Alternating Series Test says that

$$\sum \limits^\infty_{n=1} (-1)^{n-1}\,b_n$$  is convergent  if

(1) $$b_{n+1} \leq b_n\,\forall \,n$$   and

(2)  $$\lim\limits_{n\to\infty} b_n = 0$$

Test for divergence says that if  $$\lim\limits_{n\to\infty} a_n \neq 0$$ or the limit doesn't exist then  $$\sum\limits_{n=1}^\infty a_n$$ diverges .

Consider the option (a),

$$\sum\limits_{n=1}^\infty \dfrac{(-1)^{n-1}}{2n-1} = \sum\limits_{n=1}^\infty (-1)^{n-1} b_n$$

$$\therefore b_n = \dfrac{1}{2n-1} , \,\,b_{n+1} = \dfrac{1}{2n+1}$$

Clearly,  $$b_{n+1} \leq b_n\; \forall \,\,n$$

Also,  $$\lim\limits_{n\to\infty} b_n = \lim\limits_{n\to\infty} \dfrac{1}{2n-1} =0$$

$$\therefore$$ Series is convergent by Alternating Series Test.

Consider the option (b),

$$\sum\limits_{n=1}^\infty (-1)^{n-1}\,n\,e^{-n} = \sum\limits_{n=1}^\infty (-1)^{n-1}\,n\,e^{-n} = \sum\limits_{n=1}^\infty (-1)^{n-1}\,b_n$$

$$\therefore b_n = n\,e^{-n}, \,\,b_{n+1} = (n+1)\,e^{-(n+1)}$$

To test whether the series is decreasing, consider

$$f(x) = x\,e^{-x}$$

$$\Rightarrow f'(x) = e^{-x} - x\,e^{-x} >0$$

$$\Rightarrow 1-x>0$$

$$\Rightarrow x<1$$

$$\Rightarrow f$$ is increasing  for  $$x<1$$  and decreasing for $$x>1$$ .

$$\therefore b_{n+1} \leq \,b_n\;\; \forall \,\,n>1$$

Also  $$\lim\limits_{n\to\infty} b_n =\lim\limits_{n\to\infty} \dfrac{n}{e^n} =0$$         (use L' Hospital's Rule)

$$\therefore$$ Series is convergent.

Consider the option (c)

$$\sum\limits_{n=1}^\infty (-1)^{n-1}\,\dfrac{3n-1}{2n+1} =\sum\limits_{n=1}^\infty (-1)^{n-1}\, b_n$$

$$\therefore b_n = \dfrac{3n-1}{2n+1}$$

Here,  $$\lim\limits_{n\to \infty} b_{n}\, = \lim\limits_{n\to \infty} \dfrac{3n-1}{2n+1}\, = \lim\limits_{n\to \infty} \dfrac{3-\dfrac{1}{n}}{2+\dfrac{1}{n}} = \dfrac{3}{2}$$

$$\therefore$$ Nothing can be concluded for Alternating Series Test.

Now consider,

$$\lim\limits_{n\to\infty} (-1)^{n-1} \dfrac{3n-1}{2n+1} \to$$  which is non existent .

$$\therefore\;\lim\limits _{n\to\infty}a_n$$ is non existent, so the series diverges by test of divergence .

Consider the option (d),

Clearly,   (as   is an increasing function and angles are decreasing with increase in )

Also,

Series converges by Alternating Series Test.

Option (c) is correct.

Using Alternating Series Test and test for divergence, choose the divergent series from the following

A

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

.

B

$$\sum\limits_{n=1}^\infty (-1)^{n-1}\,n\,e^{-n}$$

C

$$\sum\limits_{n=1}^\infty (-1)^{n}\,\dfrac{3n-1}{2n+1}$$

D

$$\sum\limits_{n=1}^\infty (-1)^{n}\,\dfrac{sin\,\pi}{n}$$

Option C is Correct

Alternating Series Estimation Theorem

• If  $$S= \sum\limits_{n=1}^\infty (-1)^{n-1} \, b_n$$ is the sum of an alternating series that satisfies.

(a)  $$b_{n+1} \leq b_n \,\,{\text{and}}\,\,$$  (b)  $$\lim\limits_{n\to\infty} b_n =0$$

Then  $$|R_n | = |S-S_n| \leq b_{n+1}$$

The above statement is the Alternating Series Estimation Theorem .

How many terms of the given series do we need to take in order to find the sum to accuracy such that $$|error | < .00005$$ ?  $$\sum\limits_{n=1}^\infty \dfrac{(-1)^{n-1}}{n^6}$$

A $$n=5$$

B $$n=4$$

C $$n=3$$

D $$n=2$$

×

For alternating series

$$|R_n| = |error| =|S-S_n | \leq b_{n+1}$$

If  $$b_{n+1} \leq b_n \,\,{\text{ (for all n) and}}\,\, \lim\limits_{n\to\infty} \;b_n=0$$

In this case  $$\sum\limits^\infty_{n=1} \dfrac{(-1)^{n-1}}{n^6}$$

$$\Rightarrow b_n = \dfrac{1}{n^6}$$

$$\therefore |R_n| = |S-S_n| \leq \dfrac{1}{(n+1)^6}$$

Now $$|error| < .00005$$

$$\Rightarrow \dfrac{1}{(n+1)^6} < .00005$$

$$\Rightarrow (n+1)^6 > \dfrac{1}{.00005}$$

$$\Rightarrow (n+1)^6 > 20000$$

$$\Rightarrow n+1 > 5.21$$

$$\Rightarrow n>4.21$$

$$\therefore n=5 \to$$  we need to add at least 5 terms of the series.

How many terms of the given series do we need to take in order to find the sum to accuracy such that $$|error | < .00005$$ ?  $$\sum\limits_{n=1}^\infty \dfrac{(-1)^{n-1}}{n^6}$$

A

$$n=5$$

.

B

$$n=4$$

C

$$n=3$$

D

$$n=2$$

Option A is Correct

The Root Test for Convergence

(1) If  $$\lim\limits_{n\to\infty} \sqrt[n]{|a_n|} = L<1$$  then the series $$\sum\limits^\infty_{n=1} a_n$$ is absolutely convergent (and therefore convergent)

(2) If $$\lim\limits_{n\to\infty} \sqrt[n]{|a_n|} = L>1$$ or $$\infty$$ then series  $$\sum\limits^\infty_{n=1} a_n$$  is divergent.

(3) If  $$\lim\limits_{n\to\infty} \sqrt[n]{|a_n|} =1$$ then root test is inconclusive .

The Root test is convenient to apply when $$n^x$$ power occurs.

e.g. Consider the series  $$\sum\limits^\infty_{n=1} \left(\dfrac{2n+3}{3n+2}\right)^n$$

Here , $$a_n=\left(\dfrac{2n+3}{3n+2}\right)^n$$

$$\Rightarrow \sqrt[n]{|a_n|}=\left|\dfrac{2n+3}{3n+2}\right|$$

Now  $$\lim\limits_{n\to\infty} \sqrt[n]{a_n} = \lim\limits_{n\to\infty} \left(\dfrac{2n+3}{3n+2}\right) = \lim\limits_{n\to\infty} \left(\dfrac{2+\dfrac{3}{n}}{3+\dfrac{2}{n}}\right)$$

$$= \dfrac{2}{3}$$

$$\therefore$$ the series is convergent by ratio test.

Using Root test, find out which of the following series converge? (1)  $$\sum\limits_{n=1}^\infty \left(\dfrac{2n^2+3}{3n^2+4}\right)^n$$      (2)  $$\sum\limits_{n=1}^\infty \left(\dfrac{-2n}{n+1}\right)^{5n}$$

A Only 1

B Only 2

C both 1 and 2

D None

×

(1) If  $$\lim\limits_{n\to\infty} \sqrt[n]{|a_n|} = L<1$$  then the series $$\sum\limits^\infty_{n=1} a_n$$ is absolutely convergent (and therefore convergent)

(2) If $$\lim\limits_{n\to\infty} \sqrt[n]{|a_n|} = L>1$$ or $$\infty$$ then series  $$\sum\limits^\infty_{n=1} a_n$$  is divergent.

Consider the given series

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

$$\therefore a_n = \left(\dfrac{2n^2+3}{3n^2+4}\right)^n$$

$$\Rightarrow \sqrt[n]{a_n} = \dfrac{2n^2+3}{3n^2+4}$$

$$\therefore \lim\limits_{n\to\infty} \sqrt[n]{|a_n|} = \lim\limits_{n\to\infty}\dfrac{2n^2+3}{3n^2+4}= \lim\limits_{n\to\infty} \dfrac{2+\dfrac{3}{n^2}}{3+\dfrac{4}{n^2}} = \dfrac{2}{3} <1$$

$$\therefore$$ the given series is convergent by Root test.

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

$$\therefore a_n = \left(\dfrac{-2n}{n+1}\right)^{5n}$$

$$\Rightarrow \sqrt[n]{|a_n|} = \left(\dfrac{2n}{n+1}\right)^5$$

$$\therefore \; \lim\limits_{n\to\infty}\sqrt[n]{|a_n|} = \lim\limits_{n\to\infty} \left(\dfrac{2n}{n+1}\right)^5= \lim\limits_{n\to\infty} \left(\dfrac{2}{1+\dfrac{1}{n}}\right)^5$$

$$= (+2)^5 = 32 >1$$

$$\therefore$$ the given series diverges by Root test.

$$\therefore$$ Correct option is (a)

Using Root test, find out which of the following series converge? (1)  $$\sum\limits_{n=1}^\infty \left(\dfrac{2n^2+3}{3n^2+4}\right)^n$$      (2)  $$\sum\limits_{n=1}^\infty \left(\dfrac{-2n}{n+1}\right)^{5n}$$

A

Only 1

.

B

Only 2

C

both 1 and 2

D

None

Option A is Correct

Power Series

• A series of the form   $$\sum\limits_{n=0}^\infty C_n \,x^n = C_0 + C_1 \,x + C_2 x^2 +C_3 x^3 + .....$$

is called a power series, where  $$x$$ is a variable and $$C'_n s$$ are constants called the coefficients of the series.

• For a fixed $$x$$ , this series is a series of constants which we can test for convergence or divergence. It may converge for some value of $$x$$ and diverge for the other values.
• We define the sum of series as a function

$$f(x) = C_0 + C_1 x +C_2 x^2+ C_3x^3 + ........C_n x^n +.....$$

• The domain of this function 'f' is the set of all $$x$$ for which the series converge. 'f' is not a polynomial as it has infinite number of terms.
• If  $$C_n =1\;\; \,\,\forall\,\,\,n$$

$$f(x) = 1+x+x^2+x^3 +.......$$     which is a geometric series which converges for  $$|x| <1$$ and diverges for $$|x| \geq1$$  .

• In general     $$\sum\limits_{n=0}^\infty C_n (x-a)^n = C_0 +C_1 (x-a) +C_2 (x-a)^2+.....$$

is called a power series in $$(x-a)$$ or power series centered at 'a' or  power series about 'a' .

For what values of  $$x$$  will the following power series converge? $$\sum\limits_{n=0}^{\infty} \dfrac{(x-2)^n}{n^2+1}$$

A $$1<x<3$$

B $$5<x<8$$

C $$2<x<4$$

D $$x>10$$

×

Apply Ratio test for convergence

$$\Rightarrow\sum\limits_{n=0}^\infty \dfrac{(x-2)^n}{n^2+1} = \sum\limits_{n=0}^\infty a_n$$

$$\Rightarrow a_n =\dfrac{(x-2)^n}{n^2+1}$$

$$\Rightarrow \left|\dfrac{a_{n+1}}{a_n}\right| = \left|\dfrac{(x-2)^{n+1}}{\dfrac{(n+1)^2+1}{\dfrac{(x-2)^n}{n^2+1}}}\right| = |x-2|× \dfrac{n^2+1}{n^2+2n+2}$$

$$\therefore \lim\limits_{n\to\infty}\left|\dfrac{a_{n+1}}{a_n}\right| = \lim\limits_{n\to\infty}|x-2| × \dfrac{n^2+1}{n^2+2n+2}$$

$$=\lim\limits_{n\to\infty} |x-2| × \left(\dfrac{1+\dfrac{1}{n^2}}{1+\dfrac{2}{n}+\dfrac{2}{n^2}}\right) = |x-2|$$

For convergence  $$|x-2| <1$$

$$\Rightarrow -1<x-2<1$$

$$\Rightarrow 1<x<3$$

$$\Rightarrow x\in(1,3)$$

For what values of  $$x$$  will the following power series converge? $$\sum\limits_{n=0}^{\infty} \dfrac{(x-2)^n}{n^2+1}$$

A

$$1<x<3$$

.

B

$$5<x<8$$

C

$$2<x<4$$

D

$$x>10$$

Option A is Correct

Absolute and Conditional Convergence (Ratio Test)

• Given any series $$\sum\limits^\infty_{n=1} a_n$$ we consider a corresponding series

?$$\sum\limits^\infty_{n=1} |a_n | =|a_1 |+|a_2 |+.....|a_n |$$ each of whose terms are absolute value of the terms of original series.

• Series $$\sum a_n$$ is called absolutely convergent if the series  $$\sum |a_n|$$ is convergent.
• For series with positive terms, absolute convergence is same as convergence.
• Series $$\sum a_n$$ called conditionally convergent if it is convergent but not absolutely convergent.

For example, the  series

$$1-\dfrac{1}{2} + \dfrac{1}{3}-\dfrac{1}{4} +.......$$

$$= \sum\limits_{n=1}^\infty \, (-1)^{n+1} \dfrac{1}{n}$$

is conditionally convergent because

$$\sum\limits_{n=1}^\infty \,\left|\dfrac{1}{n} × (-1)^{n+1}\right| = \sum\limits_{n=1}^\infty \dfrac{1}{n}$$  is divergent

but  $$\sum\limits_{n=1}^\infty \, (-1)^{n+1} \dfrac{1}{n}$$  is convergent .

• If series $$\sum a_n$$ is absolutely convergent then it is convergent.

For example, the series

$$\sum\limits_{n=1}^\infty \dfrac{(-1)^{n+1}}{n^2}$$  is convergent because  $$\sum\limits_{n=1}^\infty \dfrac{1}{n^2}$$ is convergent.

• The ratio test is an important tool to determine whether given series is absolutely convergent.

• Ratio Test

(i) If  $$\lim\limits_{n\to \infty}|\dfrac{a_{n+1}}{a_n}| = L <1$$  then $$\sum a_n$$ is absolutely convergent (and therefore convergent also)

(ii) If $$\lim\limits_{n\to \infty}|\dfrac{a_{n+1}}{a_n}| = L >1 \,\,\text{or} \,\,=\infty$$  then $$\sum a_n$$ is divergent.

(iii) If $$\lim\limits_{n\to \infty}|\dfrac{a_{n+1}}{a_n}| = 1$$ the ratio test is inconclusive and nothing can be said about convergence and divergence of $$\sum a_n$$.

Find which of the following series are absolutely convergent (use ratio test) (1) $$\sum\limits_{n=1}^\infty \dfrac{n^{100}\,100^n}{n \,!}$$       (2) $$\sum\limits_{n=1}^\infty \dfrac{(-3)^n}{(2n+1) \,!}$$        (3) $$\sum\limits_{n=1}^\infty \dfrac{(-1)^n × (1.2)^n}{n^4 }$$

A 1 and 3 only

B 1 and 2 only

C 2 and 3 only

D 1,2 and 3 only

×

Ratio Test

(i) If  $$\lim\limits_{n\to \infty}|\dfrac{a_{n+1}}{a_n}| = L <1$$  then $$\sum a_n$$ is absolutely convergence.

(ii) If $$\lim\limits_{n\to \infty}|\dfrac{a_{n+1}}{a_n}| = L >1 \,\,\text{or} \,\,=\infty$$  then $$\sum a_n$$ is divergently.

(iii) If $$\lim\limits_{n\to \infty}|\dfrac{a_{n+1}}{a_n}| = 1$$ the n nothing can be said about convergence

Consider the option (a)

$$\sum\limits_{n=1}^\infty \dfrac{n^{100}× 100^n}{n\,!}$$

$$\Rightarrow a_n = \dfrac{n^{100}× 100^n}{n\,!}$$

$$\therefore \left|\dfrac{a_{n+1}}{a_n}\right| = \left|\dfrac{\dfrac{(n+1)^{100}× 100^{n+1}}{(n+1)!}}{\dfrac{n^{100}× 100^n}{n\,!}}\right| = \left(\dfrac{n+1}{n}\right)^{100}×\dfrac{100}{n+1}$$

$$= \left(1+\dfrac{1}{n}\right)^{100} × \dfrac{100}{n+1}$$

$$\therefore \lim\limits_{n\to\infty} \left|\dfrac{a_{n+1}}{a_n}\right| = \lim\limits_{n\to\infty} \left(1+\dfrac{1}{n}\right)^{100} × \dfrac{100}{n+1} = 1×0 =0$$

which is less than 1 $$\Rightarrow$$ series is absolutely convergent.

Consider the option (b)

$$\sum\limits_{n=1}^\infty \dfrac{(-3)^n}{(2n+1) \,!}$$

$$\Rightarrow a_n = \dfrac{(-3)^n}{(2n+1)\,!}$$

$$\therefore \left|\dfrac{a_{n+1}}{a_n}\right| = \left|\dfrac{\dfrac{(-3)^{n+1}}{(2n+3)!}}{\dfrac{(-3)^{n}}{(2n+1)!}}\right| = \left|\dfrac{-3}{(2n+2)(2n+3)}\right| = \dfrac{3}{(2n+2)(2n+3)}$$

$$\therefore \lim\limits_{n\to\infty} \left|\dfrac{a_{n+1}}{a_n}\right| = \lim\limits_{n\to \infty} \dfrac{3}{(2n+2)(2n+3)} = 0$$

which is less than 1  $$\Rightarrow$$ series is absolutely convergent .

Consider the option (c)

$$\sum\limits_{n=1}^\infty \dfrac{(-1)^n × (1.2)^n}{n^4 }$$

$$\Rightarrow a_n= \dfrac{(-1)^n × (1.2)^n}{n^4 }$$

$$\therefore \left|\dfrac{a_{n+1}}{a_n}\right| =\left |\dfrac{\dfrac{(-1)^{n+1}× (1.2)^{n+1}}{(n+1)^4}}{\dfrac{(-1)^n × (1.2)^n}{n^4}}\right| = \dfrac{1.2}{\left(\dfrac{n+1}{n}\right)^4}$$

$$= \dfrac{1.2}{\left(1+\dfrac{1}{n}\right)^4}$$

$$\therefore \lim\limits_{n\to \infty} \left|\dfrac{a_{n+1}}{a_n}\right| = \lim\limits_{n\to \infty}\dfrac{1.2}{\left(1+\dfrac{1}{n}\right)^4} =1.2$$

which is greater than 1  $$\Rightarrow$$ series is divergent.

$$\therefore$$ Correct option is (b).

Find which of the following series are absolutely convergent (use ratio test) (1) $$\sum\limits_{n=1}^\infty \dfrac{n^{100}\,100^n}{n \,!}$$       (2) $$\sum\limits_{n=1}^\infty \dfrac{(-3)^n}{(2n+1) \,!}$$        (3) $$\sum\limits_{n=1}^\infty \dfrac{(-1)^n × (1.2)^n}{n^4 }$$

A

1 and 3 only

.

B

1 and 2 only

C

2 and 3 only

D

1,2 and 3 only

Option B is Correct

The Ratio Test

• Ratio Test

(i) If  $$\lim\limits_{n\to \infty}|\dfrac{a_{n+1}}{a_n}| = L <1$$  then $$\sum a_n$$ is absolutely convergent (and therefore convergent also)

(ii) If $$\lim\limits_{n\to \infty}|\dfrac{a_{n+1}}{a_n}| = L >1 \,\,\text{or} \,\,=\infty$$  then $$\sum a_n$$ is divergent.

(iii) If $$\lim\limits_{n\to \infty}|\dfrac{a_{n+1}}{a_n}| = 1$$ the ratio test is inconclusive and nothing can be said about convergence and divergence of $$\sum a_n$$.

Choose the series for which the ratio test is inconclusive (i.e. it doesn't tell us about convergence or divergence). (1) $$\sum\limits_{n=1}^\infty \dfrac{1}{n^4}$$       (2) $$\sum\limits_{n=1}^\infty \dfrac{n}{3^n}$$      (3) $$\sum\limits_{n=1}^\infty \dfrac{(-3)^{n-1}}{\sqrt n}$$        (4) $$\sum\limits_{n=1}^\infty \dfrac{\sqrt n}{1+n^2}$$

A 1 and 2

B 2 and 3

C 1 and 4

D 3 and 4

×

Ratio test is inconclusive about convergence or divergence of a series if

$$\lim\limits_{n\to \infty} \left|\dfrac{a_{n+1}}{a_n}\right| = 1$$

Consider the given series

(1)     $$\sum\limits_{n=1}^\infty \dfrac{1}{n^4} \to a_n = \dfrac{1}{n^4}$$

$$\therefore \left|\dfrac{a_{n+1}}{a_n}\right| = \left|\dfrac{n^4}{(n+1)^4}\right| = \left|\left(\dfrac{n}{n+1}\right)^4\right| = \left(\dfrac{1}{1+\dfrac{1}{n}}\right)^4$$

$$\therefore \lim\limits_{n\to\infty} \left|\dfrac{a_{n+1}}{a_n}\right|= \lim\limits_{n\to\infty} \left(\dfrac{1}{1+\dfrac{1}{n}}\right)^4 =1$$

$$\therefore$$ Ratio test is inconclusive .

(2)   $$\sum\limits_{n=1}^\infty \dfrac{n}{3^n} \to a_n = \dfrac{n}{3^n}$$

$$\therefore \left|\dfrac{a_{n+1}}{a_n}\right| =\left|\dfrac{n+1× 3^n}{3^{n+1}× n}\right| = \dfrac{n+1}{3n} = \dfrac{1}{3}\left(1+\dfrac{1}{n}\right)$$

$$\therefore \lim\limits_{n\to \infty} \left|\dfrac{a_{n+1}}{a_n}\right| = \lim\limits_{n\to\infty}\dfrac{1}{3} \left(1+\dfrac{1}{n}\right) = \dfrac{1}{3}<1$$

$$\therefore$$ series converges

(3)   $$\sum\limits_{n=1}^\infty \dfrac{(-3)^{n-1}}{\sqrt n} \to a_n = \dfrac{(-3)^{n-1}}{\sqrt n}$$

$$\therefore \left|\dfrac{a_{n+1}}{a_n}\right| =\left|\dfrac{(-3)^n}{\sqrt{n+1}×\dfrac{(-3)^{n-1}}{\sqrt n}}\right| = \left|3\sqrt{\dfrac{n}{n+1}}\right| =\left|3\sqrt{\dfrac{1}{1+\dfrac{1}{n}}}\right|$$

$$\therefore \lim\limits_{n\to\infty}\; 3\;\sqrt{\dfrac{1}{1+\dfrac{1}{n}}} = 3>1$$

$$\therefore$$ series diverges .

(4)    $$\sum\limits_{n=1}^\infty \dfrac{\sqrt n}{1+n^2} \to a_n = \dfrac{\sqrt n}{1+n^2}$$

$$\therefore\left|\dfrac{a_{n+1}}{a_n}\right| = \left|\dfrac{\dfrac{\sqrt{n+1}}{1+(n+1)^2}}{\dfrac{\sqrt n}{1+n^2}}\right| = \left|\dfrac{\sqrt{n+1}}{\sqrt n} × \dfrac{n^2+1}{n^2+2n+2}\right|$$

$$= \sqrt{1+\dfrac{1}{n}} × \dfrac{1+\dfrac{1}{n^2}}{1+\dfrac{2}{n}+\dfrac{2}{n^2}}$$

$$\lim\limits_{n\to\infty} \sqrt{1+\dfrac{1}{n}} × \dfrac{1+\dfrac{1}{n^2}}{1+\dfrac{2}{n}+\dfrac{2}{n^2}} = 1× 1=1$$

$$\therefore$$ ratio test is inconclusive .

$$\therefore$$ Option (c) is correct.

Choose the series for which the ratio test is inconclusive (i.e. it doesn't tell us about convergence or divergence). (1) $$\sum\limits_{n=1}^\infty \dfrac{1}{n^4}$$       (2) $$\sum\limits_{n=1}^\infty \dfrac{n}{3^n}$$      (3) $$\sum\limits_{n=1}^\infty \dfrac{(-3)^{n-1}}{\sqrt n}$$        (4) $$\sum\limits_{n=1}^\infty \dfrac{\sqrt n}{1+n^2}$$

A

1 and 2

.

B

2 and 3

C

1 and 4

D

3 and 4

Option C is Correct