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Infinite Series Formula helps to calculate the sum of a sequence where the number of terms is infinite. Infinite series is the series that contains infinite terms and Sn which is the sum of the first n terms is known as the partial sum of the given infinite series. If Sn is given a limit as n moves towards infinity that limit is known as the sum to infinity of the series and the result is termed as the sum of infinity of series.
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Key Takeaways: Infinite series formula, sum of infinite geometric series, geometric progression, series infinity
Infinite Series
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The arithmetic series is a sequence in which the difference between each consecutive term remains constant throughout, whereas the geometric series is a sequence in which the ratio of consecutive terms to the preceding term remains constant throughout.
A sequence a1 , a2 , a3 ,…, an ,… is called arithmetic sequence or arithmetic progression if an + 1 = an + d, n ∈ N, where a1 is known as the first term and the constant term d is referred to as the common difference of the A.P or Arithmetic Progression.
Some properties of Arithmetic Progression are:
- When each term of an A.P. is added up by a constant, the resulting sequence is also an A.P.
- If you subtract a constant from each term of an A.P. The resulting sequence is also an A.P.
- If an Arithmetic Progression's terms are multiplied by a constant, the resulting sequence is also an A.P.
- If an Arithmetic Progression's terms are divided by a non-zero constant, the resulting sequence is also an A.P.
If the series has infinite terms, it is referred to as an infinite series, and the sum of the first n terms, Sn, is referred to as a partial sum of the given infinite series. If the partial sum, i.e. the sum of the first n terms, Sn, is given a limit as n approaches infinity, the limit is referred to as the sum to infinity of the series, and the result is referred to as the sum of the infinite series.
Sum of Infinite Series Formula
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To find the sum of a series that extends to infinity, you can use the sum of the infinite geometric series formula. This is also referred to as the sum of infinite Geometric Progression.
While calculating the sum of a GP, we discover that the sum converges to a value, despite the fact that the series has infinite terms. If the infinite series formula -1<r<1, could be given as, S\(\infty\)= \(\frac{a_1}{1-r}\)
Here,
a= first term of the series, and r= a common ratio between two consecutive terms and -1<r<1
If d > 0, the sum of an infinite arithmetic sequence is \(\infty\)
If d > 0- \(\infty\), d > 0, the sum of an infinite arithmetic sequence is \(\infty\)
Relationship between A.M. and G.M.
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Let A and G be A.M. and G.M. of two given positive real numbers a and b, respectively.
Then A=\(\frac{a + b}{2}\) and G = \(\sqrt{ab}\)
Thus, we have A – G =\(\frac{a + b}{2}\) – \(\sqrt{ab}\) = \(\frac{a + b- 2\sqrt{ab}}{2}\)
= \(\frac{(\sqrt{a} - \sqrt{b})^2}{2}\)\( \geq\) 0
Therefore, we obtain the relationship A ≥ G.
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Things to Remember
- For an arithmetic series, the sum of infinite is undefined because the sum of terms leads to .
- If r is greater than one, the sum does not exist because it does not converge.
- If |r|< 1, the sum of a geometric series can be calculated to infinity.
- a= first term of the series, and r= a common ratio between two consecutive terms
- The formula for the sum of infinite series is,
S\(\infty\)= \(\frac{a}{1-r}\) or \(\lim\limits_{n} \to \)\(\infty\)Sn = S\(\infty\)= \(\frac{a}{1-r}\)
Sample Questions
Ques: With the use of the infinite series formula, find the sum to infinite series: ½ + 1/6 +1/18 + 1/54+ …. (3 marks)
Ans: We have, the first term, a=½ and,
the common ratio, r= \(\frac{\frac{1}{6}}{\frac{1}{2}}\)= \(\frac{1}{3}\)
In order to find the sum of given infinite series, if r<1, the sum is given as, S= \(\frac{a}{1-r}\)
Hence applying the values to the infinite series formula, to find
Sum = \(\frac{\frac{1}{2}}{1-\frac{1}{3}}\)
Sum= \(\frac{\frac{1}{2}}{\frac{2}{3}}\)
thus, Sum = \(\frac{3}{4}\)
Thus, the sum of ½+ 1/6+ 1/18+ 1/54+ ……is \(\frac{3}{4}\) .
Ques: If exists, for the series 24+12+6+, find the sum to infinity? (3 marks)
Ans: We have, 64+ 16+4+...
Where, a= 64, and common ratio is, 16/64 = ¼
To find the sum to infinity, we can r=¼ < 1
Sum = 
Therefore, the sum to infinite terms of the given series is \(\frac{256}{3}\).
Ques: Evaluate the equation : \(\displaystyle\sum_{0}^{\infty}(\frac{1}{2})^n\) (2 marks)
Ans. The sum of the given series is, ![]()
Here, the first term, a=1 and the common term, r=½
We know Infinite series formula is, S\(\infty\)= \(\frac{a}{1-r}\)
So, S\(\infty\)= \(\frac{1}{1-\frac{1}{2}}\)= \(\frac{1}{\frac{1}{2}}\)= 2
Ques: Evaluate the equation : 3+7+11+........ (2 marks)
Ans: we have first term, a=3, d=4, and n= \(\infty\)
Here, the difference is larger than 0,
Hence, the sum = +\(\infty\)
Ques: Find the sum of the Infinite series : ¼+1/16+1/64+1/256+.... , using the infinite series formula. (3 marks)
Answer: Given, a=¼
r = \(\frac{\frac{1}{16}}{\frac{1}{4}}\)=\(\frac{4}{16}\)= \(\frac{1}{4}\)
To find the sum to the given infinite series, if r<1, is then the sum is given as S\(\infty\)= \(\frac{a}{1-r}\)
Hence, when we apply the values to the infinite series formula, we get,
Sum= \(\frac{\frac{1}{4}}{1-\frac{1}{4}}\)
Sum = \(\frac{\frac{1}{4}}{\frac{3}{4}}\)
Sum = \(\frac{4}{12}\), Sum = \(\frac{1}{3}\)
Therefore, the sum to the infinite series ¼+1/16+1/64/1/256+..... is \(\frac{1}{3}\).
Q6: find the sum to n terms of the series 5+11+19+29+41….. (4 marks)
Ans: Here, Sn=

Ques: Find the sum to n terms of the series whose nth term is n (n+3). (3 marks)
Ans: We have, an=n (n+3)=n2+3n
Thus, the sum to n terms is given by,
![]()
Therefore,
= ![]()
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