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Sum and series
Definitions
Let a1 , a2 , ... ai , ... be a sequence of numbers, complex or real. Then we may consider
the infinite series, (or simply series):
∞
X
ai = a1 + a2 + a3 ...
(1)
i=1
The ai are called the terms of the series. The sum of the first n terms is:
sn =
n
X
ai = a1 + a2 + ... + an
(2)
i=1
This expression is called the nth partial sum of the series (1). Clearly, if we omit the
terms of sn from (1), the remaining expression is:
Rn =
∞
X
ai = an+1 + an+2 + an+3 + ...
(3)
i=n+1
The is called the remainder of the series (1) after the nth term.
In this way, we have now associated with the series (1) the sequence if its partial sums
s1 , s2 , s3 , ... If the sequence is convergent, say,
lim sn = s
n→∞
(4)
then the series (1) is said convergent, the number s is called its value (or sum), and
we write:
∞
X
s=
ai = a1 + a2 + a3 + ...
(5)
i=1
If the sequence of the partial sums diverges, the series (1) is said to divergent.
If the series (1) converges and has the value s then
s = sn + Rn
(6)
From the definition of the convergence of a sequence it follows that |Rn | can be made as
small as we want, by taking n large enough. In many cases it will be impossible to find
the sum s of a convergent series. Then for computational purposes we must use a partial
sum sn as an approximation of s, and by estimating the remainder Rn we may obtain
information about the accuracy of the approximation.
Examples
The series
∞
X
1 1 1
1
= + + + ...
i
2
2 4 8
i=1
(7)
converges and has value 1 because
1 1
1
1
1
sn = + + ... + n = 1 − n and lim 1 − n = 1
n→∞
2 4
2
2
2
The series
∞
X
(8)
i = 1 + 2 + 3 + ...
(9)
(−1)i = 1 − 1 + 1 − 1 + ...
(10)
i=1
diverges.
The series
∞
X
i=0
diverges because:
s0 = 1 , s1 = 0 , s2 = 1 , s3 = 0 , ...
(11)
and the sequence 0, 1, 0, 1, ... is divergent.
The harmonic series
∞
X
1
i=1
i
=1+
1 1 1
+ + + ...
2 3 4
diverges. In fact
(12)
1
1
+ ... +
(13)
2
n
and sn equals the sum of the areas of the n rectangles (see figure here below). This area
is greater than the area An under the corresponding portion of the curve y = 1/x, and
Z n+1
dx
An =
= ln(n + 1) → ∞ as n → ∞
(14)
x
1
sn = 1 +
Since sn > An , it follows that sn → ∞ as n → ∞, which means divergence.
2
1.8
1.6
y = 1/X
1.4
1.2
1
0.8
0.6
0.4
0.2
0
1
2
3
4
x
2
...
n
n+1
Some useful series
1
1 1 1
+ + +
+ ... = 2
2 4 8 16
1 1 1 1
1 + + + + + ... diverges
2 3 4 5
1 1 1 1
1 − + − + − ... = ln 2
2 3 4 5
1
1
π2
1 1
+
+ ... =
1+ + +
4 9 16 25
6
π
1 1 1 1
1 − + − + − ... =
3 5 7 9
4
1+
(15)
(16)
(17)
(18)
(19)
(20)
A useful sequences
n(a1 + an )
2
if ai+1 − ai = constant ∀i
sn = a1 + a2 + ... + an =
(21)
(22)
Particular cases:
1 + 2 + 3 + 4 + ... + N =
N(N + 1)
2
(23)
1 + 3 + 5 + 7 + ... + N =
(1 + N)2
4
(24)
An exotic series (due to Ramanujan)
∞
X
(4k)!(1103 + 26390k)
k=0
(k!)4 3964k
3
9801
= √
2 2π
(25)