Download Convergence and Divergence of Series: Geometric and Harmonic Series and more Study notes Signals and Systems in PDF only on Docsity!
4.2. Series
A series is an infinite sum: ∑^ ∞
n=
an = a 1 + a 2 + a 3 + · · · + an + · · ·
In order to define the value of this sum we start be defining its sequence of partial sums
sn =
∑^ n
i=
ai = a 1 + a 2 + · · · + an.
Then, if limn→∞ sn = s exists the series is called convergent and its sum is that limit: ∑∞
n=
an = s = lim n→∞ sn.
Otherwise the series is called divergent.
For instance, consider the following series: 1 2
2 n^
∑^ ∞
n=
2 n^
Its partial sums are:
sn =
∑^ n
i=
2 i^
2 n^
2 n^
Hence its sum is ∑^ ∞
n=
2 n^
= lim n→∞
∑^ n
i=
2 i^
= lim n→∞
2 n
4.2.1. Geometric Series. A series verifying an+1 = ran, where r is a constant, is called geometric series. If the first term is a 6 = 0 then the series is
a + ar + ar^2 + · · · + arn^ + · · · =
∑^ ∞
n=
arn^.
The partial sums are now:
sn =
∑^ n
i=
ari^.
The nth partial sum can be found in the following way:
sn = a + ar + ar^2 + · · · + arn rsn = ar + ar^2 + · · · + arn^ + arn+
hence
sn − rsn = a + 0 + 0 + · · · + 0 − arn+1^ ,
so:
sn =
a(1 − rn+1) 1 − r
If |r| < 1 we can rewrite the result like this:
sn =
a 1 − r
a 1 − r
rn+1^ ,
and then get the limit as n → ∞:
s = lim n→∞
sn =
a 1 − r
a 1 − r
lim n→∞
rn+ ︸ ︷︷ ︸ ↓ 0
a 1 − r
So for |r| < 1 the series is convergent and
∑^ ∞
n=
arn^ =
a 1 − r
For |r| ≥ 1 the series is divergent.
4.2.2. Telescopic Series. A telescopic series is a series whose terms can be rewritten so that most of them cancel out.
Example: Find
∑^ ∞
n=
n(n + 1)
Answer : Note that
n(n + 1)
n
n + 1
. So the nth partial sum
is
sn =
∑^ ∞
i=
i
i + 1
n
n + 1 = 1 −
n + 1
Example: Show that
∑^ ∞
n=
sin n diverges.
Answer : All we need to show is that sin n does not tend to 0. If for some value of n, sin n ≈ 0, then n ≈ kπ for some integer k, but then
sin (n + 1) = sin n cos 1 + cos n sin 1 ≈ sin kπ cos 1 + cos kπ sin 1 = 0 ± sin 1 = ± 0. 84 · · · 6 = 0
So if a term sin n is close to zero, the next term sin (n + 1) will be far from zero, so it is impossible for sin n to get permanently closer and closer to 0.
4.2.6. Operations with Series. If
n=1 an^ and^
n=1 bn^ are con- vergent series and c is a constant then the following series are also convergent and:
∑^ ∞
n=
can = c
∑^ ∞
n=
an
∑^ ∞
n=
(an + bn) =
∑^ ∞
n=
an +
∑^ ∞
n=
bn
∑^ ∞
n=
(an − bn) =
∑^ ∞
n=
an −
∑^ ∞
n=
bn
