Summation of Series

Arithmetic-Geometric Progression

A sequence in which every term is a product of a term of AP and GP is know as arithmetic-geometric progression.

The series may be written as

$a,\left(a+d\right)r,\left(a+2d\right)r^2,\left(a+3d\right)r^3,.....,\left[a+\left(n-1\right)d\right]r^{n-1}$

$S_n=\frac{a}{1-r}+\frac{dr\left(1-r^{n-1}\right)}{{\left(1-r\right)}^2}-\frac{\left\{a+\left(n-1\right)d\right\}{r}^n}{1-r},ifr\ne 1$

$S_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right],ifr=1$

$S_{\infty }=\frac{a}{1-r}+\frac{dr}{{\left(1-r\right)}^2},if\left|r\right|<1$

Sum of Arithmetic-Geometric Series

Type 1. Let $a_1+a_2+a_3+.........$be a given series, if $a_2-a_1,a_3-a_2.......$are in AP or GP, then $a_n$ and $S_n$ can be found by the method of difference.

Let $S_n=a_1+a_2+a_3+a_4+.....+a_n$

$S_n=a_1+a_2+a_3+.......+a_{n-1}+a_n$

So, $S_n-S_n=a_1+\left(a_2-a_1\right)+\left(a_3-a_2\right)+\left(a_4-a_3\right)+\left(a_n-a_{n-1}\right)-a_n$

$a_n=a_1\left(a_2-a_1\right)+\left(a_3-a_2\right)+......+\left(a_n-a_{n-1}\right)$

$a_n=a_1+T_1+T_2+T_3+.....+T_{n-1}$

Where, $T_1+T_2+T_3+.....$ are terms of new series and $S_n=\sum a_n$

Type 2. It is not always necessary that the series of first order of differences i.e. $a_2-a_1,a_3-a_2,....,a_n-a_{n-1}$ is always either in AP or in GP in such case.

Let $a_1=T_1,a_2-a_1=T_2,a_3-a_2=T_3,....,a_n-a_{n-1}=T_n$

So, $a_n=T_1+T_2+.......+T_n$

$a_n=T_1+T_2+,......+T_{n-1}+T_n$

On subtracting above two equations , we get

$T_n=T_1+\left(T_2-T_1\right)+\left(T_3-T_2\right)+......+\left(T_n-T_{n-1}\right)$

Now, The series $\left(T_2-T_1\right)+\left(T_3-T_2\right)+......+\left(T_n-T_{n-1}\right)$is series of second

Order of differences and when it is either in AP or in GP, then $a_n=a_1+\sum T_r.$

Otherwise, in the similar way, we find series of higher order of differences and the $n$ith term of the series,

Exponential Series

The sum of the series $1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+....\infty$ is denoted by the number $e.$

$e=\underset{x\to \infty }{\lim }{\left(1+\frac{1}{n}\right)}^n$

$\therefore e=1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+.....=\underset{x\to \infty }{\lim }{\left(1+\frac{1}{n}\right)}^n$

1 $e$ lies between 2and 3.

2 $e$ is an irrational number.

3 $e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+......\infty$

4 $e^{-x}=1-\frac{x}{1!}+\frac{x^2}{2!}-\frac{x^3}{3!}+.....\infty$

Exponential Theorem

Let, $a>0,$ then for all real values of $x$,

$a^x=1+x\left({\log }_{e}a\right)+\frac{x^2}{2!}{\left({\log }_{e}a\right)}^2+\frac{x^3}{3!}{\left({\log }_{e}a\right)}^3+........\infty$

Logarithmic Series

1. $\begin{array}{r}{\log }_e\left(1+x\right)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+......\infty \\ \therefore \log \left(1+x\right)=\sum _{n=1}^{\infty }{\left(-1\right)}^{n-1}\frac{x^n}{n}\\ \end{array}$

2. $\begin{array}{r}{\log }_e\left(1-x\right)=-x-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}-......\infty \\ \Rightarrow -{\log }_e\left(1-x\right)=x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+.....\infty \\ \end{array}$

3. ${\log }_e\left(\frac{1+x}{1-x}\right)=2\left(x+\frac{x^3}{3}+\frac{x^5}{5}+....\infty \right)$

4. ${\log }_{e}2=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-......\infty$

Some Key Series and Results

1 $\sum _{n=0}^{\infty }\frac{1}{n!}=e=\sum _{n=0}^{\infty }\frac{1}{\left(n-1\right)!}=\sum _{n=0}^{\infty }\frac{1}{\left(n-k\right)!}=e$

2 $\sum _{n=0}^{\infty }\frac{1}{n!}=\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+......\infty =e-1$

3 $\sum _{n=2}^{\infty }\frac{1}{n!}=\frac{1}{2!}+\frac{1}{3!}+.....\infty =e-2$

4 $\sum _{n=0}^{\infty }\frac{1}{\left(n+1\right)!}=\frac{1}{1!}+\frac{1}{2!}+\frac{1}{4!}+.....\infty =e-1$

5 $\sum _{n=1}^{\infty }\frac{1}{\left(n+1\right)!}=\sum _{n=0}^{\infty }\frac{1}{\left(n+2\right)!}=\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+......\infty =e-2$

6 $\sum _{n=0}^{\infty }\frac{1}{\left(2n\right)!}=1+\frac{1}{2!}+\frac{1}{4!}+\frac{1}{6!}+.....=\frac{e+e^{-1}}{2}=\sum _{n=1}^{\infty }\frac{1}{\left(2n-2\right)!}$

7 $\sum _{n=1}^{\infty }\frac{1}{\left(2n-1\right)!}=\frac{1}{1!}+\frac{1}{3!}+\frac{1}{5!}+.....=\frac{e-e^{-1}}{2}=\sum _{n=0}^{\infty }\frac{1}{\left(2n+1\right)!}$

8 $e^{ax}=1+\frac{\left(ax\right)}{1!}+\frac{{\left(ax\right)}^2}{2!}+\frac{{\left(ax\right)}^3}{3!}+.......+\frac{{\left(ax\right)}^n}{n!}+....\infty$

9 $\sum _{n=0}^{\infty }\frac{n}{n!}=e=\sum _{n=1}^{\infty }\frac{n}{n!}$

10 $\sum _{n=0}^{\infty }\frac{n^2}{n!}=2e=\sum _{n=1}^{\infty }\frac{n^2}{n!}$

11 $\sum _{n=0}^{\infty }\frac{n^3}{n!}=5e=\sum _{n=1}^{\infty }\frac{n^3}{n!}$

12 $\sum _{n=0}^{\infty }\frac{n^4}{n!}=15e=\sum _{n=1}^{\infty }\frac{n^4}{n!}$

13 $\sum _{r=1}^n\left(a_r\pm b_r\right)=\sum _{r=1}^n{a}_r\pm \sum _{r=1}^n{b}_r$

14 $\sum _{r=1}^{n}k{a}_r=k\sum _{r=1}^n{a}_r$

15 $\sum _{r=1}^{n}k=k+k....n$ times $n.k,$ where $k$ is a constant.

16 $\sum _{r=1}^{n}r=1+2+.....+n=\frac{n\left(n+1\right)}{2}$

17 $\sum _{r=1}^n{r}^2=1^2+2^2+3^3+.....+n^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}$

18 $\sum _{r=1}^n{r}^3=1^3+2^3+3^3+.....+n^3={\left[\frac{n\left(n+1\right)}{2}\right]}^2$

19 $\sum _{r=1}^n{r}^4=1^4+2^4+3^3+n^4=\frac{n\left(n+1\right)\left(6n^3+9n^2+n-1\right)}{30}$

20 $2\sum _{i<j=1}^n{a}_i{a}_j={\left(a_1+a_2+.....+a_n\right)}^2-\left(a_1^2+a_2^2+a_3^2+....+a_n^2\right)$

21 Sum of first $n$ even natural numbers.

i.e. $2+4+6+.....+2n=n\left(n+1\right)$

22. Sum of first $n$ odd natural numbers.

i.e. $1+3+5+.......+\left(2n-1\right)=n^2$

23. Sum of $n$ terms of series

$1^2-2^2+3^2-4^2+5^2-6^2+7^2-8^2+....$

Case I When $n$ is odd $=\frac{n\left(n+1\right)}{2}$

Case II When $n$ is even $=\frac{-n\left(n+1\right)}{2}$

24. If number of terms in AP/GP/HP are odd, then AM/GM/HM of first and last term is middle term of progression.

25. If $p$th, $q$th and $r$th term of geometric progression are also in progression , then $p,q$and $r$ are also in geometric progression .

26. If $a,b$ and $c$ are in AP and also in GP, then $a=b=c$

27. If $a,b,$ and $c$ are in AP, then $x^a,x^b$and $x^c$ are in geometric progression