After having studied the limits of number sequences, we can apply our newly acquired knowledge to topics which occur more often in physics, for instance infinite sums s =
∞
P
n=1
an, calledseries:
These are often encountered sometimes in more interesting physical questions: For in- stance if we want to sum up the electrostatic energy of infinitely many equidistant alter- nating positive and negative point charges for one chain link (which gives a simple but surprisingly good one-dimensional model of a ion crystal) we come across the infinite sum over the members of the alternating harmonic sequence (F7): the series
∞
P
n=1
(−1)n+1 n . How do we calculate this?
Series are sequences whose members are finite sums of real numbers: The definition of a
series
∞
P
n=1
an assequence of partial sums sm = m
P
n=1
an
m∈N
reduces the series to sequences which we have been dealing with just above.
Especially, a series is exactly then convergentand has the value s, if the sequence of its partial sumssm (not that of itssummands an!!) converges: lim
m→∞sm =s:
series sm =
m
P
n=1
an convergent ⇐⇒ lim
m→∞
m
P
n=1
an=s <∞
Also the multiple of a convergent series and the sum and difference of two convergent series are again convergent.
The fewsampleseries that we need, to see the most important concepts, we derive simply through piecewise summing up our sample sequences:
(R1) The series of the partial sums of the sequence (F1) of the natural numbers:
sm = m
P
n=1
n
m∈N
= 1,3,6,10,15, . . . is clearly divergent.
(R2)The series made out of the members of the alternating sequence (F2) always jumps between 1 and 0 and has therefore two cluster points and consequently no limit.
(R3)Also the “harmonic series” summed up out of the members of the harmonic sequence (F3), i.e. the sequence
sm =
m
P
n=1 1 n
m∈N
= 1, 32, 116, 2512, 13760, . . .is divergent. Because the (also necessary) Cauchy Criterion is not fulfilled: If we for instance chooseε = 14 >0 and consider a piece of the sequence forn = 2mconsisting ofmterms: |s2m−sm|=
2m
P
n=m+1 1 n =
1
m+1 + m+21 +. . .+ 2m1 > 1 2m + 1
2m +. . .+ 1 2m
| {z }
m summands
= 12 > ε = 14 while for convergence < ε would have been necessary.
(R7) Their alternating variant however, created out of the sequence (F7), our physical example from above, converges
∞
P
n=1
(−1)n+1
n (= ln 2, as we will show later).
Because of this difference between series with purely positive summands and alternating ones, it is appropriate to introduce a new term: A series is said to beabsolutely convergent, if already the series of the absolute values converges.
Series sm =
m
P
n=1
an absolutely convergent ⇐⇒ lim
m→∞
m
P
n=1
|an|<∞
We can easily understand that within an absolutely convergent series the summands can be rearranged without any effect on the limiting value. Two absolutely convergent series can be multiplied termwise to create a new absolutely convergent series.
For absolute convergence the mathematicians have developed various sufficient criteria, the so-calledmajorant criteria which you will deal with more closely in the lecture about analysis:
Insert: Majorants: If a convergent majorant sequence S= lim
m→∞Sm =
∞
P
n=1
Mn
exists with positive Mn > 0, whose members are larger than the corresponding absolute values of the sequence under examination Mn ≥ |an|, then the series
m→∞lim sm =
∞
P
n=1
an is absolutely convergent, because from the Triangle Inequality it follows
|sm|=|
m
P
n=1
an| ≤
m
P
n=1
|an| ≤
m
P
n=1
Mn=Sm. Very often the “geometric series”
(R6):
∞
P
n=0
qn, which follow from the geometric sequences (F6) (qn)n∈N, q ∈ R , serve as majorants. To calculate them we benefit from the earlier forq6= 1 derived geometric sum:
m→∞lim
m
X
n=0
qn = lim
m→∞
1−qm+1
1−q = 1
1−q <∞, meaning convergent for|q|<1 and divergent for |q| ≥1.
Insert: Quotient criterion: We present here as example for a majorant criterion only the quotient criterion which is obtained through comparison with the geometric series:
If lim
n→∞|an+1a
n |<1, issm=
m
P
n=1
an absolutely convergent.
As an example we prove the absolute convergence of the series (R9)
∞
P
n=0
nqn for
|q|<1, which can be obtained from the for |q| < 1 convergent geometric series (R6) through termwise multiplication with the divergent sequence (F1) of the natural numbers. We calculate therefore
n→∞lim
an+1
an
= lim
n→∞
(n+ 1)qn+1 nqn
=|q| lim
n→∞
n+ 1
n =|q|<1.
That the criterion is not necessary can be seen from the series (R8), the summing up of the sample sequence (F8):
∞
P
n=1 1
n2 = π62, which is absolutely convergent, since all members are positive, but
n→∞lim
n2
(n+1)2 = lim
n→∞
1
(1+n−1)2 = 1.
(R4) The series of the inverse natural factorials
∞
P
n=1 1
n! deserves to be examined in more detail:
First we realize that the sequence of the partial sums
sm =
m
P
n=1 1 n!
m∈N
increases mono- tonically: sm+1−sm = (m+1)!+1 > 0. To get an upper bound B we estimate through the majorant geometric sum with q= 12:
|sm| = 1 + 1 2!+ 1
3!+. . .+ 1 m!
< 1 + 1 2 + 1
22 +. . .+ 1 2m−1
=
m−1
X
n=0
(1 2)n
= 1−(12)m 1− 12
< 1
1− 12 = 2.
Since the monotonically increasing sequence of the partial sumssm is bounded from above by B = 2 the Theorem of Bolzano and Weierstrass guarantees us convergence. We just do not know the limiting value yet. This limit is indeed something fully new - namely an irrational number. We call it =e−1, so that the number e after the supplementary convention 0! = 1 is defined by the following series starting withn= 0:
Exponential series defined by: e:=
∞
P
n=o 1 n!.
Insert: The number e is irrational: we prove indirectly that the so defined number e is irrational, meaning it cannot be presented as quotient of two integers g and h:
If ewere writable in the form e= gh with integersg and h≥2, then h!e= (h−1)!g would be an integer:
However, from definition it holds (h−1)!g = h!e=h!
∞
X
n=0
1 n! =
h
X
n=0
h!
n!+
∞
X
n=h+1
h!
n!
=
h! +h! +h!
2!+ h!
3! +. . .+ 1
+ + lim
n→∞
1
h+ 1+ 1
(h+ 1)(h+ 2) +. . .+ 1
(h+ 1)(h+ 2). . .(h+n)
.
While the first bracket is an integer if h is, this cannot be true for the second bracket, because
1
h+ 1 + 1
(h+ 1)(h+ 2) +. . .+ 1
(h+ 1)(h+ 2). . .(h+n) =...
= 1
h+ 1
1 + 1
h+ 2+. . .+ 1
(h+ 2). . .(h+n)
,
which can be estimated through the geometric series with q= 12 as follows,
< 1 h+ 1
1 +1
2 +. . .+ 1 2n−1
= 1
h+ 1·1−(12)n 1−(12)
< 1
h+ 1· 1
1−(12) = 2
h+ 1 ≤2/3,
Because h should be h ≥ 2 there is a contradiction. Consequently e must be irra- tional.
To get the numerical value of e we first calculate the members of the zero sequence (F4) an= n!1:
a1 = 1!1 = 1, a2 = 2!1 = 12 = 0.50, a3 = 3!1 = 16 = 0.1666, a4 = 4!1 = 241 = 0.041 666, a5 = 5!1 = 1201 = 0.008 33, a6 = 6!1 = 7201 = 0.001 388, a7 = 7!1 = 5 0401 = 0.000 198, a8 = 8!1 = 40 3201 = 0.000 024, a9 = 9!1 = 362 8801 = 0.000 002, . . . then we sum up the partial sums: sm =
m
P
n=1 1
n! = 1 + 2!1 +3!1 +4!1 +. . .+ m!1 s1 = 1, s2 = 1.50, s3 = 1.666 666, s4 = 1.708 333,
s5 = 1.716 666, s6 = 1.718 055, s7 = 1.718 253, s8 = 1.718 278, s9 = 1.718 281, . . ..
If we look at the rapid convergence, we can easily imagine that after a short calculation we receive the following result for the limiting value: e= 2.718 281 828 459 045. . .
Insert: A sequence converging to e: Besides this exponential series which we used to define e there exists as earlier mentioned in addition a sequence, con- verging to the number e, the exponential sequence(F10):
(1 + 1n)n
n∈N= 2, (32)2, (43)3, . . . ,which we will shortly deal with for comparison:
According to the binomial formula we find firstly for the general sequence member:
an = (1 + 1 n)n=
n
X
k=0
n!
(n−k)!k!nk
= 1 +n
n+n(n−1)
n22! +n(n−1)(n−2)
n33! +. . .+n(n−1)(n−2). . .(n−(k−1))
nkk! +
. . .+ n!
nnn!
= 1 + 1 +(1−n1)
2! + (1−n1)(1−n2)
3! +. . .+ (1−n1)(1−n2). . .(1−k−1n )
k! +. . .
+(1−n1)(1− 2n). . .(1−n−1n ) n!
On the one hand we enlargethis expression foran, by forgetting the subtraction of the multiples of 1n within the brackets:
an≤1 + 1 + 1 2!+ 1
3!+. . .+ 1
n! = 1 +sn
and reach so (besides the term one) the corresponding partial sums of the exponential series sn. Thus the exponential series is a majorant for the also monotonically increasing exponential sequence and ensures the convergence of the sequence through that of the series. For the limiting value we get:
n→∞lim an≤e.
On the other hand we diminish the above expression for an by keeping only the first (k+ 1) of the without exception positive summands and throwing away the other ones:
an≥1 + 1 +(1−1n)
2! +(1−n1)(1−2n)
3! +. . .+(1− 1n)(1−2n). . .(1− (k−1)n )
k! .
When we now first let the larger n, of the two natural numbers tend to infinity, we get:
a:= lim
n→∞an≥1 + 1 + 1 2! + 1
3!+. . .+ 1
k! = 1 +sk
and after letting also the smaller natural number k tend to infinity we reach:
a≥e.
Consequently the limit a:= lim
n→∞an of the exponential sequence an must be equal to the number e defined by the exponential series:
n→∞lim(1 +n1)n=
∞
P
n=0 1 n! =e
When you, however, calculate the members of the sequence and compare them with the partial sums of the series, you will realize that the sequence converges much more slowly than the series.
Through these considerations we now have got a first overview over the limiting procedures and some of the sequences and series important for natural sciences with their limits, which will be of great use for us in the future.