Unit-I: Sets and Functions
Unit-II: Algebra
Unit-III: Coordinate Geometry
Unit-IV: Calculus
Unit-V: Mathematical Reasoning
Unit-VI: Statistics and Probability

Arithmetico-geometric series

Suppose  (a_1,a_2,...,a_n,...) \text { F} (a_n)  is an A.P and  (b_n)=(b_1, b_2,...b_n,...) is in G.P, then  (a_nb_n) is known as an Arithmetico-geometric sequence.

Accordingly  \Sigma \ a_nb_n is called an Arithmetic-geometric series.

An Arithmetic-geometric sequence is of the form,  ab, (a+d)br, (a+2d)br^{2}, (a+3d)br^{3},...

[where  d and  r are respectively common difference and common ratio corresponding to the arithmetic series  \Sigma a_n and the geometric series  \Sigma b_n ].

Partial Sum

Sum of  n terms of an Arithmetic-geometric series is,

 S_n=ab\dfrac{1-r^n}{1-r}+db\left\{ \dfrac{r}{(1-r)^{2}}-\dfrac{r^n}{(1-r)^{2}}+\dfrac{r^n}{1-r}-\dfrac{nr^n}{1-r} \right\}


 S_n=\dfrac{ab}{1-r}+\dfrac{dbr(1-r^{n-1})}{(1-r)^{2}}-\dfrac{\left[ a+(n-1)d \right] br^n}{1-r}

For  \left| r \right| <  1 i.e  -1 <  r <  1 , then  \lim_{n \to \infty } r^n=0 and  \lim_{n \to \infty } nr^n=0 .

Therefore, the sum of an infinite number of terms of the sequence is,

 \lim_{n \to \infty } S_n=\dfrac{ab}{1-r}+\dfrac{dbr}{(1-r)^{2}}

For which the series  \sum_{n=0}^{\infty }r^n is convergent.

Some Different Kinds of Series -:

(i)  Binomial Series -:

 We know from binomial theorem for positive integral index   n , that,


 +\dfrac{n(n-1)(n-2)...2.1}{n!}\ x^n .

The above expression holds for any value of  x .

Binomial Theorem for a Real Index

 (1+n)^\alpha =1+\alpha x+\dfrac{\alpha (\alpha -1)}{2!}\ x^{2}+...+\dfrac{\alpha (\alpha -1) ...(\alpha -n+1)}{n!}\ x^n+...

where  \alpha \in R , provided |x| <  1 .

This is known as the Binomial Series and it has sum  (1+x)^\alpha  for  |x| <  1 .

 \to  If  \alpha =n\in N , the binomial series becomes the finite series and has the sum  (1+x)^n without any restriction on  x .

Series Expansion of a Function -:

If a series  t_0+t_1+t_2+... has a finite sum  's' we say that the series is convergent and write  S= t_0+t_1+t_2+...

If  f(x) is the sum of a series, the series is said to be an expansion of  f(x) .

As  1+\alpha n+\dfrac{\alpha (\alpha -1)}{2!}x^{2}... is an expansion of  (1+x)^\alpha

Application of Binomial Series

(i)   (1-x)^{-1}=1+x+x^{2}+x^{3}+... ,for |x| <  1

(ii)  (1+x)^{-1}=1-x+x^{2}-x^{3}+... , for  |x|  < 1

(iii)  (1+x)^{-2}=1-2x+3x^{2}-4x^{3}+... , for  |x| <  1

(iv)  (1-x)^{-2}=1+2x+3x^{2}+4x^{3}+... , for  |x|  < 1

(v)  \sqrt{1+x}=1+\dfrac{x}{2}-\dfrac{x^{2}}{2\cdot4}+\dfrac{1\cdot 3}{2\cdot 4\cdot 6}\ x^{3}-... , for  \left| x \right|\leq 1

(vi)  \sqrt{1-x}=1+\dfrac{x}{2}+\dfrac{x^{2}}{2\cdot4}+\dfrac{1\cdot 3}{2\cdot 4\cdot 6}\ x^{3}-... , for  \left| x \right|\leq 1

(vii)  \sqrt[3]{1+x}=1+\dfrac{x}{3}-\frac{1\cdot 2}{3\cdot 6}\ x^{2}+\dfrac{1\cdot 2\cdot 5}{3\cdot 6\cdot 9}\ x^{3}-... , for  \left| x \right|\leq 1

(viii)  \sqrt[3]{1-x}=1+\dfrac{x}{3}+\dfrac{1\cdot 2}{3\cdot 6}\ x^{2}+\dfrac{1\cdot 2\cdot 5}{3\cdot 6\cdot 9}\ x^{3}+... , for  \left| x \right|\leq 1

The Exponential Series -:

We know,  \lim_{n \to \infty } \left(1+\dfrac{1}{n}  \right)^n=e,\  n\in N

If  x is any real number, by the binomial theorem for real index we get, for  n > 1 .

 \begin{aligned} \left( 1+\dfrac{1}{n} \right)^{nx}&=1+nx\cdot \dfrac{1}{n}+ \dfrac{nx(nx-1)}{2!}\cdot \dfrac{1}{n^{2}}+...\\&=1+x+ \dfrac{x(x-\dfrac{1}{n})}{2!}+ \dfrac{n(n-\dfrac{1}{n})(x-\dfrac{2}{n})}{3!}+... \end{aligned}

Since  (1+\dfrac{1}{n}^{nx}=\left\{ \left( 1+\dfrac{1}{n} \right)^n \right\}^x , taking the limit as   , we get  e^x=1+\dfrac{x}{1!}+\dfrac{x^{2}}{2!}+\dfrac{x^{3}}{3!}+... , for  x\in R

This series is called Exponential Series and also called the expansion of  e^x .

Let  a > 0 and  b+\log_e^a=l_na

i.e.  e^b=a

 \implies e^{bx}=1+b_x+\dfrac{b^{2}x^{2}}{2!}+\dfrac{b^{2}x^{3}}{3!}+...

 \implies (Putting  bx in place of  x )

 \implies a^n=(e^b)^x=1+bx+\dfrac{b^{2}n^{2}}{2!}+\dfrac{b^{2}n^{3}}{3!}+...

 \implies a^n=1+x \log_e^a+\dfrac{x^{2}(log_e^a)^{2}}{2!}+\dfrac{x^3(log_e^a)^3}{3!}+...


 a^x=1+x\ln a+\dfrac{x^{2}(\ln a)^{2}}{2!}+\dfrac{x^{3}(\ln a)^{3}}{3!}+...

This series is also called the Exponential series and called the expansion of  a^x .

The Logarithmic Series -:

 a^y=1+y\ln a+\dfrac{y^{2}(\ln a)^{2}}{2!}+\dfrac{y^{3}\ln a)^{3}}{3!}+...

Let  a=1+x > 0 so that  x > -1 ,then

 (1+n)^y=1+y\ln(1+n)+\dfrac{y^{2}}{2!}\left\{ \ln(1+x) \right\}^{2}+...

Restricting  x so that  |x|  <  1, then


From above equations, we have  \ln(1+x)=x-\dfrac{x^{2}}{2}+\dfrac{x^{3}}{3}-..., |x| <  1.

This series is called the Logarithmic series and also called the expansion of  \ln(1+x) .

 \ln(1-x)=-x-\dfrac{x^{2}}{2}-\dfrac{x^{3}}{3}-..., \ -1\le x < 1

 \implies \ln\left( \dfrac{1+n}{1-n} \right)=2\left( x+\dfrac{x^{3}}{3}+\frac{x^{5}}{5} +...\right),\ |x| <  1.

Above two expressions are the expansion of  \ln(1-x) and  \ln\left( \dfrac{1+y}{1-x} \right) .

Some Finite Series -: (For  n terms)

(i)  1+2+3+...+n= \dfrac{n(n+1)}{2}

(ii)  2+4+6+...+2n= n(n+1)

(iii)  1+3+5+...+(2n-1)= n^{2}

(iv)  k+(k+1)+(k+2)+...+(k+n-1)= \dfrac{n(2k+n-1)}{2}

(v)  1^{2}+2^{2}+3^{2}+...+n^{2}=\dfrac{n(n+1)(2n+1)}{6}

(vi)  1^{3}+2^{3}+3^{3}+...+n^{3}=\left[ \dfrac{n(n+1)}{2} \right]^{2}

(vii. 1^{2}+3^{2}+5^{2}+...+(2n-1)^{2}=\dfrac{n(4n^{2}-1)}{3}

(viii)  1^{3}+3^{3}+5^{3}+...+(2n-1)^{3}=n^{2}(2n^{2}-1)

Some Infinite Series -:

(i)  1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{2^n}+...=2

(ii)  \dfrac{1}{1\cdot 2}+\dfrac{1}{2\cdot 3}+\dfrac{1}{3\cdot 4}+...+\dfrac{1}{n(n+1)}+...+1

(iii)  1+\dfrac{1}{1!}+\dfrac{1}{2!}+\dfrac{1}{3!}+...+\dfrac{1}{(n-1)!}+\dfrac{1}{n!}+...=e

Convergence of Infinite Series -:

 \sum_{n=1}^{\infty }a_+n=L , if  \lim_{n \to \infty }\  S_n=L

Properties of Convergent Series -:

Convergent Series -:  \sum_{n=1}^{\infty}a_n=A,\ \sum_{n=1}^{\infty }b_n=B

(i)  \sum_{n=1}^{\infty }(a_n+b_n)=\sum_{n=1}^{\infty }a_n +\sum_{n=1}^{\infty }b_n=A+B

(ii)  \sum_{n=1}^{\infty }ca_n=c\sum_{n=1}^{\infty }a_n=cA, c\in R

Convergence Tests -:

Let  \sum_{n=1}^{\infty }a_n and  \sum_{n=1}^{\infty }b_n be series such that   a_n and  b_n are positive for all  n ,i.e.  0 <  a_n  \leq b_n.

(i) If  \sum_{n=1}^{\infty }b_n is convergent, then  \sum_{n=1}^{\infty }a_n is also convergent.

(ii) If  \sum_{n=1}^{\infty }a_n is divergent, then  \sum_{n=1}^{\infty }b_n is also divergent.

The Limit Comparison Test -:

Let \sum_{n=1}^{\infty} a_n  and \sum_{n=1}^{\infty} b_n  be series such that    a_n and  b_n are positive for all  n .

(i) If  0  <  \lim_{n \to \infty} \dfrac{a_n}{b_n} then  \sum_{n=1}^{\infty }a_n and  \sum_{n=1}^{\infty }b_n are either both convergent or both divergent.

(ii) If  \lim_{n \to \infty } \dfrac{a_n}{b_n}=0 , then  \sum_{n=1}^{\infty }b_n is convergent implies that  \sum_{n=1}^{\infty }a_n is also convergent.

(iii) If  \lim_{n \to \infty } \dfrac{a_n}{b_n}=\infty  , then  \sum_{n=1}^{\infty }b_n is divergent implies that  \sum_{n=1}^{\infty }a_n is also divergent.

Absolute Convergent -:

 \to  A series  \sum_{n=1}^{\infty }a_n is absolutely convergent if the series  \sum_{n=1}^{\infty }\left| a_n \right| is convergent.

 \to  If the series series  \sum_{n=1}^{\infty }a_n is absolutely convergent then it is convergent.

Conditional Convergent -:

A series  \sum_{n=1}^{\infty }a_n is conditionally convergent if the series is convergent but it is not absolutely convergent.

The Ratio Test -:

Let  \sum_{n=1}^{\infty }a_n be a series with positive terms.

(i) If  \lim_{n \to \infty } \dfrac{a_{n+1}}{a_n} <   1 , then  \sum_{n=1}^{\infty }a_n is convergent.

(ii) If  \lim_{n \to \infty } \dfrac{a_{n+1}}{a_n} > 1 , then  \sum_{n=1}^{\infty }a_n is divergent.

(iii) If  \lim_{n \to \infty } \dfrac{a_{n+1}}{a_n}= 1 , then  \sum_{n=1}^{\infty }a_n may converge or diverge and the ratio test is inconclusive.

The Root Test -:

Let  \sum_{n=1}^{\infty }a_n be a series with positive terms.

(i) If  \lim_{n \to \infty } \sqrt[n]{a_n}\lt 1 , then  \sum_{n=1}^{\infty }a_n is convergent.

(ii) If  \lim_{n \to \infty } \sqrt[n]{a_n}\gt 1 , then  \sum_{n=1}^{\infty }a_n is divergent.

(iii) If  \lim_{n \to \infty } \sqrt[n]{a_n}= 1 , then  \sum_{n=1}^{\infty }a_n may converge or diverge, but no conclusion can be drawn from this test.

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