Series: Types, Summation Formulas, Properties, Applications & Examples 

1.0Introduction to Series

A series in mathematics is the sum of the terms of a sequence. While a sequence lists numbers in a specific order, a series focuses on their sum. In JEE Maths, series play a critical role in algebra, calculus, and problem-solving.

For example, the sum of the sequence 2, 4, 6, 8 is the series:
( 2 + 4 + 6 + 8 = 20 ).

Series can be finite (with a fixed number of terms) or infinite (continuing indefinitely). Mastery of series helps in evaluating complex expressions, solving calculus problems, and tackling various JEE questions efficiently.

2.0Types of Series

Arithmetic Series

An arithmetic series is the sum of the terms in an arithmetic progression (AP), where each term increases by a constant difference.

General Form:

$(S_n=a+(a+d)+(a+2d)+...+[a+(n-1)d])$

Sum of n Terms:

$S_n=\frac{n}{2}[2a+(n-1)d]$
or

$S_n=\frac{n}{2}(a_1+a_n)$

Example:
Sum of first 5 terms of 3, 7, 11, ...
( a = 3, d = 4, n = 5 )

$(S_5=\frac{5}{2}[2\cdot 3+(5-1)\cdot 4]=\frac{5}{2}[6+16]=\frac{5}{2}\cdot 22=55)$

Geometric Series

A geometric series is the sum of a geometric progression (GP), where each term is multiplied by a common ratio.

General Form:

$(S_n=a+ar+a{r}^2+...+a{r}^{n-1})$

Sum of n Terms $(for(r\ne 1))$:

$S_n=a\frac{r^n-1}{r-1}$

Sum to Infinity (for ( |r| < 1 )):

$S_{\infty }=\frac{a}{1-r}$

Example:
( a = 2, r = 3, n = 4 )

$(S_4=2\frac{3^4-1}{3-1}=2\frac{81-1}{2}=2\cdot 40=80)$

Harmonic Series

A harmonic series is formed by taking the sum of reciprocals of an arithmetic progression.

General Form:

$(S_n=\frac{1}{a}+\frac{1}{a+d}+\frac{1}{a+2d}+...+\frac{1}{a+(n-1)d})$

The harmonic series diverges as $(n\to \infty )$, i.e., its sum grows without bound.

Example:
The classic harmonic series: $(1+\frac{1}{2}+\frac{1}{3}+...)$

Infinite Series

• A series with an infinite number of terms.
• Examples:

$\text{Convergent: }\sum \frac{1}{2^n}=1$

$\text{Divergent: }\sum \frac{1}{n}\text{ (Harmonic series)}$

Special Series

i. Series of Natural Numbers:

$(1+2+3+...+n=\frac{n(n+1)}{2})$

ii. Series of Squares:

$(1^2+2^2+3^2+...+n^2=\frac{n(n+1)(2n+1)}{6})$

iii. Series of Cubes:

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

iv. Telescoping Series:
A series where many terms cancel each other when expanded, simplifying the sum.

3.0Summation of Series

Summation Notation

The symbol $(\sum )(\sigma )$ represents the sum of a sequence.

General Notation:

$({\sum }_{k=1}^n{a}_k=a_1+a_2+\cdots +a_n)$

Example:

$({\sum }_{k=1}^{5}k=1+2+3+4+5=15)$

Common Summation Formulas

• $({\sum }_{k=1}^{n}k=\frac{n(n+1)}{2})$
• $({\sum }_{k=1}^n{k}^2=\frac{n(n+1)(2n+1)}{6})$
• $({\sum }_{k=1}^n{k}^3={\left[\frac{n(n+1)}{2}\right]}^2)$
• $({\sum }_{k=0}^{n-1}a{r}^k=a\frac{r^n-1}{r-1})(GP)$

4.0Properties of Series

• Linearity:

$(\sum (a_k+b_k)=\sum a_k+\sum b_k)$

$(\sum c\cdot a_k=c\sum a_k)$

• Splitting:
   A series can often be split into simpler parts for easier calculation.
• Convergence/Divergence:
   For infinite series, convergence determines if the sum approaches a finite value.
• Telescoping:
   A telescoping series collapses to just a few terms after cancellation.

5.0Sum of Series: Formulas and Methods

Knowing how to sum different types of series is essential for JEE.

• Arithmetic Series:

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

• Geometric Series:

$[S_n=a\frac{1-r^n}{1-r}]$

• Sum of First ( n ) Natural Numbers:

$[1+2+3+\cdots +n=\frac{n(n+1)}{2}]$

• Sum of Squares:

$[1^2+2^2+3^2+\cdots +n^2=\frac{n(n+1)(2n+1)}{6}]$

• Sum of Cubes:

$[1^3+2^3+3^3+\cdots +n^3={\left[\frac{n(n+1)}{2}\right]}^2]$

• Telescoping Series:

If $(S=(a_1-a_2)+(a_2-a_3)+\cdots +(a_n-an+1))$,
Then $(S=a_1-an+1)$.

6.0Solved Examples on Series

Example 1: Arithmetic Series

Q: Find the sum of the first 20 terms of the series 5, 8, 11, ...

A:
( a = 5, d = 3, n = 20 )

$(S_{20}=\frac{20}{2}[2\cdot 5+(20-1)\cdot 3]=10[10+57]=10\times 67=670)$

Example 2: Geometric Series

Q: Find the sum of first 6 terms of GP: 3, 6, 12, 24, ...

A:
( a = 3, r = 2, n = 6 )

$(S_6=3\frac{2^6-1}{2-1}=3\times (64-1)=3\times 63=189)$

Example 3: Series of Squares

Q: Calculate $(1^2+2^2+3^2+...+10^2)$.

A:

$(S=\frac{10\cdot 11\cdot 21}{6}=\frac{2310}{6}=385)$

Example 4: Infinite Geometric Series

Q: Find the sum of the infinite series: $(6+2+\frac{2}{3}+...)$

A:

$(a=6,r=\frac{1}{3})$

$(S_{\infty }=\frac{6}{1-\frac{1}{3}}=\frac{6}{\frac{2}{3}}=9)$

Example 5: Telescoping Series

Q: Evaluate $({\sum }_{k=1}^{10}\frac{1}{k(k+1)})$.

A: $(\frac{1}{k(k+1)}=\frac{1}{k}-\frac{1}{k+1})$
So,

$S=(1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+...+(\frac{1}{10}-\frac{1}{11})$
All intermediate terms cancel, leaving:

$1-\frac{1}{11}=\frac{10}{11}$