A series is what you get when the terms of a sequence are added. The whole topic is a set of formulae for doing that addition without writing out the terms.
Sigma notation
∑ r = 1 n u r = u 1 + u 2 + ⋯ + u n \sum_{r=1}^{n} u_r = u_1 + u_2 + \cdots + u_n r = 1 ∑ n u r = u 1 + u 2 + ⋯ + u n
The letter under the sigma is the counter , and the numbers below and above give its first and last values.
∑ r = 1 5 2 r = 2 + 4 + 6 + 8 + 10 = 30 \sum_{r=1}^{5} 2r = 2 + 4 + 6 + 8 + 10 = 30 r = 1 ∑ 5 2 r = 2 + 4 + 6 + 8 + 10 = 30
Two properties used constantly:
∑ ( a r ± b r ) = ∑ a r ± ∑ b r ∑ k a r = k ∑ a r \sum (a_r \pm b_r) = \sum a_r \pm \sum b_r \qquad \sum ka_r = k\sum a_r ∑ ( a r ± b r ) = ∑ a r ± ∑ b r ∑ k a r = k ∑ a r
A constant term summed n n n n n n
∑ r = 1 n 5 = 5 n \sum_{r=1}^{n} 5 = 5n r = 1 ∑ n 5 = 5 n
Common mistake ∑ r = 1 n 5 \sum_{r=1}^{n} 5 ∑ r = 1 n 5 5 n 5n 5 n n n n 5 n 5n 5 n r r r
Arithmetic series
S n = n 2 [ 2 a + ( n − 1 ) d ] or S n = n 2 ( a + l ) S_n = \frac{n}{2}\big[2a + (n-1)d\big] \qquad \text{or} \qquad S_n = \frac{n}{2}(a + l) S n = 2 n [ 2 a + ( n − 1 ) d ] or S n = 2 n ( a + l )
where l l l
Find the sum of the first 20 terms of 3 , 7 , 11 , … 3, 7, 11, \ldots 3 , 7 , 11 , …
a = 3 a = 3 a = 3 d = 4 d = 4 d = 4 n = 20 n = 20 n = 20
S 20 = 20 2 [ 6 + 19 ( 4 ) ] = 10 ( 6 + 76 ) = 820 S_{20} = \frac{20}{2}\big[6 + 19(4)\big] = 10(6 + 76) = 820 S 20 = 2 20 [ 6 + 19 ( 4 ) ] = 10 ( 6 + 76 ) = 820
Check with the other form: the 20th term is 79 79 79 20 2 ( 3 + 79 ) = 10 ( 82 ) = 820 \frac{20}{2}(3 + 79) = 10(82) = 820 2 20 ( 3 + 79 ) = 10 ( 82 ) = 820
Exam tip The second formula is really "average term times number of terms": a + l 2 \frac{a+l}{2} 2 a + l
Geometric series
S n = a ( 1 − r n ) 1 − r ( r ≠ 1 ) S_n = \frac{a(1-r^n)}{1-r} \qquad (r \ne 1) S n = 1 − r a ( 1 − r n ) ( r = 1 )
Find the sum of the first 6 terms of 2 , 6 , 18 , … 2, 6, 18, \ldots 2 , 6 , 18 , …
a = 2 a = 2 a = 2 r = 3 r = 3 r = 3 n = 6 n = 6 n = 6
S 6 = 2 ( 1 − 3 6 ) 1 − 3 = 2 ( 1 − 729 ) − 2 = − 1456 − 2 = 728 S_6 = \frac{2\left(1 - 3^6\right)}{1 - 3} = \frac{2(1 - 729)}{-2} = \frac{-1456}{-2} = 728 S 6 = 1 − 3 2 ( 1 − 3 6 ) = − 2 2 ( 1 − 729 ) = − 2 − 1456 = 728
Check by adding: 2 + 6 + 18 + 54 + 162 + 486 = 728 2 + 6 + 18 + 54 + 162 + 486 = 728 2 + 6 + 18 + 54 + 162 + 486 = 728
Exam tip When r > 1 r > 1 r > 1 a ( r n − 1 ) r − 1 \dfrac{a(r^n - 1)}{r-1} r − 1 a ( r n − 1 ) − 1 -1 − 1
Sum to infinity
If ∣ r ∣ < 1 |r| < 1 ∣ r ∣ < 1
S ∞ = a 1 − r S_\infty = \frac{a}{1-r} S ∞ = 1 − r a
Find 8 + 4 + 2 + 1 + ⋯ 8 + 4 + 2 + 1 + \cdots 8 + 4 + 2 + 1 + ⋯
a = 8 a = 8 a = 8 r = 1 2 r = \frac12 r = 2 1
S ∞ = 8 1 − 1 2 = 8 1 2 = 16 S_\infty = \frac{8}{1 - \frac12} = \frac{8}{\frac12} = 16 S ∞ = 1 − 2 1 8 = 2 1 8 = 16
The partial sums run 8 , 12 , 14 , 15 , 15.5 , … 8, 12, 14, 15, 15.5, \ldots 8 , 12 , 14 , 15 , 15.5 , …
Key result A sum to infinity exists only when ∣ r ∣ < 1 |r| < 1 ∣ r ∣ < 1 ∣ r ∣ ≥ 1 |r| \ge 1 ∣ r ∣ ≥ 1 diverges — the formula gives a number, but that number means nothing.
Common mistake With r = 2 r = 2 r = 2 a − 1 = − a \frac{a}{-1} = -a − 1 a = − a ∣ r ∣ < 1 |r| < 1 ∣ r ∣ < 1
Standard summation results
∑ r = 1 n r = n ( n + 1 ) 2 \sum_{r=1}^{n} r = \frac{n(n+1)}{2} r = 1 ∑ n r = 2 n ( n + 1 ) ∑ r = 1 n r 2 = n ( n + 1 ) ( 2 n + 1 ) 6 \sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6} r = 1 ∑ n r 2 = 6 n ( n + 1 ) ( 2 n + 1 ) ∑ r = 1 n r 3 = [ n ( n + 1 ) 2 ] 2 \sum_{r=1}^{n} r^3 = \left[\frac{n(n+1)}{2}\right]^2 r = 1 ∑ n r 3 = [ 2 n ( n + 1 ) ] 2
Note the third is the square of the first — a genuinely surprising fact.
∑ r = 1 10 r 2 = 10 ( 11 ) ( 21 ) 6 = 2310 6 = 385 \sum_{r=1}^{10} r^2 = \frac{10(11)(21)}{6} = \frac{2310}{6} = 385 r = 1 ∑ 10 r 2 = 6 10 ( 11 ) ( 21 ) = 6 2310 = 385
Combining them. Find ∑ r = 1 n ( 2 r + 1 ) \displaystyle\sum_{r=1}^{n}(2r+1) r = 1 ∑ n ( 2 r + 1 )
= 2 ∑ r = 1 n r + ∑ r = 1 n 1 = 2 ⋅ n ( n + 1 ) 2 + n = n ( n + 1 ) + n = n 2 + 2 n = n ( n + 2 ) = 2\sum_{r=1}^{n} r + \sum_{r=1}^{n} 1 = 2 \cdot \frac{n(n+1)}{2} + n = n(n+1) + n = n^2 + 2n = n(n+2) = 2 r = 1 ∑ n r + r = 1 ∑ n 1 = 2 ⋅ 2 n ( n + 1 ) + n = n ( n + 1 ) + n = n 2 + 2 n = n ( n + 2 )
Check at n = 3 n = 3 n = 3 3 + 5 + 7 = 15 3 + 5 + 7 = 15 3 + 5 + 7 = 15 3 ( 5 ) = 15 3(5) = 15 3 ( 5 ) = 15
Exam tip Always test a derived formula on a small case you can add by hand. It takes ten seconds and catches nearly every algebraic slip.
Sums not starting at 1
∑ r = m n = ∑ r = 1 n − ∑ r = 1 m − 1 \sum_{r=m}^{n} = \sum_{r=1}^{n} - \sum_{r=1}^{m-1} r = m ∑ n = r = 1 ∑ n − r = 1 ∑ m − 1
So ∑ r = 5 10 r = 55 − 10 = 45 \sum_{r=5}^{10} r = 55 - 10 = 45 ∑ r = 5 10 r = 55 − 10 = 45
The method of differences
When each term can be written as a difference f ( r ) − f ( r + 1 ) f(r) - f(r+1) f ( r ) − f ( r + 1 )
∑ r = 1 n ( 1 r − 1 r + 1 ) \sum_{r=1}^{n}\left(\frac1r - \frac{1}{r+1}\right) r = 1 ∑ n ( r 1 − r + 1 1 )
Writing out the terms:
( 1 1 − 1 2 ) + ( 1 2 − 1 3 ) + ( 1 3 − 1 4 ) + ⋯ + ( 1 n − 1 n + 1 ) \left(\frac11 - \frac12\right) + \left(\frac12 - \frac13\right) + \left(\frac13 - \frac14\right) + \cdots + \left(\frac1n - \frac{1}{n+1}\right) ( 1 1 − 2 1 ) + ( 2 1 − 3 1 ) + ( 3 1 − 4 1 ) + ⋯ + ( n 1 − n + 1 1 )
Every interior term appears twice with opposite signs, so the sum telescopes to
1 − 1 n + 1 = n n + 1 1 - \frac{1}{n+1} = \frac{n}{n+1} 1 − n + 1 1 = n + 1 n
Since 1 r − 1 r + 1 = 1 r ( r + 1 ) \frac1r - \frac{1}{r+1} = \frac{1}{r(r+1)} r 1 − r + 1 1 = r ( r + 1 ) 1
∑ r = 1 n 1 r ( r + 1 ) = n n + 1 \sum_{r=1}^{n}\frac{1}{r(r+1)} = \frac{n}{n+1} r = 1 ∑ n r ( r + 1 ) 1 = n + 1 n
Check at n = 3 n = 3 n = 3 1 2 + 1 6 + 1 12 = 6 + 2 + 1 12 = 9 12 = 3 4 \frac12 + \frac16 + \frac{1}{12} = \frac{6+2+1}{12} = \frac{9}{12} = \frac34 2 1 + 6 1 + 12 1 = 12 6 + 2 + 1 = 12 9 = 4 3 3 4 \frac{3}{4} 4 3
As n → ∞ n \to \infty n → ∞
Key result Write out the first three terms and the last two. The pattern of cancellation becomes obvious, and what survives is always a few terms from each end.
Common mistake Questions often give you the fraction and expect you to produce the difference yourself — by partial fractions. 1 r ( r + 1 ) \frac{1}{r(r+1)} r ( r + 1 ) 1 1 r − 1 r + 1 \frac1r - \frac{1}{r+1} r 1 − r + 1 1