Schinazi R.B. From Calculus to Analysis
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120 4 Fifty Ways to Estimate the Number π
(b) Take p in (0, 1). Show that the series
∞
n=1
(2n)!
(n!)
2
p
n
(1 −p)
n
converges for all p =1/2.
14. In this exercise we show that Stirling’s estimate is actually a lower bound of n!.
Let
u
n
=1 +(n −1/2) ln
n −1
n
and S
n
=
n
k=2
u
k
.
(a) Show that for n ≥2, we have
ln(1 −1/n) < −
1
n
−
1
2n
2
−
1
3n
3
.
(b) Use (a) to show that u
n
< 0 for all n ≥2.
(c) Show that S
n
is a decreasing sequence.
(d) Conclude that for all n ≥2,
n!>
√
2πe
−n
n
n+1/2
.
4.3 Convergence of Infinite Products
In this section we are interested in the convergence of sequences of the type
q
n
=
n
i=1
a
i
.
In the preceding section Wallis’ formula is such an infinite product. As the reader
will see, there is a close connection to infinite series. We first show that the only
interesting infinite products are the ones whose general term converges to 1.
Example 4.1 Let a
n
be a sequence such that a
n
=0 for all n ≥1. Let q
n
=
n
i=1
a
i
.
Assume that q
n
converges to a nonzero limit . This implies that a
n
converges to 1.
The proof is very easy. Note that
a
n
=
q
n
q
n−1
.
Since q
n
converges to =0, we have that
q
n
q
n−1
converges to 1 (why?). Thus, a
n
converges to 1, and we are done.
4.3 Convergence of Infinite Products 121
We are now going to find necessary and sufficient conditions for the convergence
of infinite products in two particular cases. First, we consider the case where the
general term of the product is larger than 1.
Convergence of
∞
i=1
(1 +u
i
)
Assume that u
i
is a sequence of positive numbers. The sequence
q
n
=
n
i=1
(1 +u
i
)
converges if and only if the series
∞
i=1
u
i
converges.
At first glance the reader may be surprised that an infinite product of reals strictly
larger than 1 may converge to a finite limit. What the result above says is that conver-
gence happens if only if the general term of the product converges to 1 fast enough.
Assume first that the infinite series
∞
i=1
u
i
converges. Recall that
exp(x) =
∞
n=0
x
n
n!
≥1 +x
for x ≥0. Thus,
q
n
=
n
i=1
(1 +u
i
) ≤
n
i=1
exp(u
i
) =exp
n
i=1
u
i
≤exp
∞
i=1
u
i
=C,
where the last inequality comes from the facts that the sequence u
i
is assumed to be
positive and exp is an increasing function. Note also that since
∞
i=1
u
i
converges,
the constant C is finite. On the other hand,
q
n+1
=(1 +u
n+1
)q
n
≥q
n
.
That is, q
n
is an increasing sequence. Thus, q
n
is increasing and bounded, and thus
it must converge. We have proved one implication.
For the converse, assume that q
n
converges. Let
S
n
=
n
i=1
u
i
.
Then S
n
is increasing since u
i
≥0. Note that
S
n
=1 +u
1
+···+u
n
≤(1 +u
1
)(1 +u
2
) ···(1 +u
n
) =
n
i=1
(1 +u
i
) =q
n
.
122 4 Fifty Ways to Estimate the Number π
Since q
n
converges, it must be bounded. Therefore, S
n
is also bounded, and since
it is increasing, it converges. That is, the infinite series
∞
i=1
u
i
converges, and the
proof is complete.
Example 4.2 Consider the Wallis’ product
n
i=1
4i
2
4i
2
−1
. We already know that it
converges to π/2. If we did not know that, can we apply the criterion above to
check that it converges?
Note that
4i
2
4i
2
−1
=
4i
2
−1
4i
2
−1
+
1
4i
2
−1
=1 +
1
4i
2
−1
=1 +u
i
,
where
u
i
=
1
4i
2
−1
> 0.
The criterion above may be applied since the general term of the product is of the
type 1 +u
i
with u
i
≥0. Note that
lim
n→∞
1
4i
2
−1
1
4i
2
=1.
By the p test the series
∞
i=1
1
4i
2
converges. By the limit comparison test
∞
i=1
1
4i
2
−1
converges as well. According to our criterion,
n
i=1
4i
2
4i
2
−1
converges, and we are done.
We now turn to another particular case of product. This time we assume that the
general term is less than 1.
Convergence of
∞
i=1
(1 −u
i
)
Assume that u
i
is a sequence of numbers in [0, 1). The sequence
q
n
=
n
i=1
(1 −u
i
)
converges to a strictly positive limit if and only if the series
∞
i=1
u
i
converges.
4.3 Convergence of Infinite Products 123
It may be surprising that an infinite product of numbers strictly less than 1 con-
verges to a nonzero limit. Similarly to the result above, this happens if and only if
the general term of the product converges to 1 fast enough.
We first show that q
n
converges. Note that q
n
> 0 and that
q
n+1
=(1 −u
n+1
)q
n
≤q
n
.
Thus, q
n
is decreasing and bounded below by 0, and thus it must converge to some
limit . Moreover, since q
n
≤1 for all n ≥1, we must have ≤1. Is =0? This is
a more delicate point.
Assume first that the series
∞
i=1
u
i
converges. Define the sequence
S
n
=
n
i=1
u
i
.
By the definition of convergence of the series
∞
i=1
u
i
, the sequence S
n
converges
to the limit S =
∞
i=1
u
i
. Set =1/2. There is N such that if n ≥N , then
|S
n
−S|< 1/2.
Thus,
∞
i=N +1
u
i
=
∞
i=N +1
u
i
< 1/2.
We now show, for all naturals k, that
N+k
i=N +1
(1 −u
i
) ≥1 −
N+k
i=N +1
u
i
.
We first verify that the inequality holds for k =1:
1 −u
N+1
=1 −u
N+1
.
Therefore, the inequality holds for k =1. Assume that it holds for k. Then,
N+k+1
i=N +1
(1 −u
i
) =(1 −u
N+k+1
)
N+k
i=N +1
(1 −u
i
) ≥(1 −u
N+k+1
)
1 −
N+k
i=N +1
u
i
.
Expanding the last term yields
(1 −u
N+k+1
)
1 −
N+k
i=N +1
u
i
=1 −
N+k
i=N +1
u
i
−u
N+k+1
+u
N+k+1
N+k
i=N +1
u
i
≥1 −
N+k+1
i=N +1
u
i
.
This proves the inequality by induction. Putting together the inequality just proved
and that
∞
i=N +1
u
i
< 1/2,
124 4 Fifty Ways to Estimate the Number π
we get, for all k ≥1,
N+k
i=N +1
(1 −u
i
) ≥1 −
N+k
i=N +1
u
i
≥1 −
∞
i=N +1
u
i
> 1 −1/2 =1/2.
That is, for all n>N,wehave
n
i=N +1
(1 −u
i
)>1/2.
Observe now that, for n>N,
q
n
=q
N
n
i=N +1
(1 −u
i
)>
1
2
q
N
.
In particular, the limit is larger than
1
2
q
N
> 0. This proves that >0.
We now deal with the converse. Assume that >0. There exists an integer K
such that if k ≥K, then
|q
k
−|</2.
In particular,
q
k
>/2fork ≥K.
On the other hand, it is easy to show that
1 −x ≤e
−x
for all x ≥0.
For if g is defined by g(x) =exp(−x)−(1−x), then g is differentiable, and g
(x) =
−exp(−x)+1 ≥0forx ≥0 (why?). Thus, g is increasing on [0, ∞), and so g(x) ≥
g(0) =0 for all x ≥0. This proves the inequality. Therefore,
q
k
=
k
i=1
(1 −u
i
) ≤
k
i=1
exp(−u
i
) =exp
−
k
i=1
u
i
=exp(−S
k
).
Since q
k
>/2fork ≥K,wehave
ln
exp(−S
k
)
≥ln(q
k
)>ln(/2) for k ≥K.
Hence,
S
k
< −ln(/2) for k ≥K.
Therefore, S
n
is bounded above (why?) and increasing (why?). Thus, S
n
converges,
and we are done.
Example 4.3 Does the sequence
q
n
=
n
i=1
cos
π/2
i+1
4.3 Convergence of Infinite Products 125
converge? We have
cos
π/2
i+1
=1 −u
i
with u
i
= 1 −cos(π/2
i+1
) ≥ 0. Therefore, this sequence converges. Does it con-
verge to a strictly positive limit?
Recall that
cosx ≥1 −x
2
/2 for all x.
Thus,
0 ≤u
i
≤
π/2
i+1
2
/2 =
π
2
2
2i+3
.
The series
∞
i=1
π
2
2
2i+3
=
π
2
8
1
3
converges. Thus, the series
∞
i=1
u
i
converges as well (why?). Therefore, the infi-
nite product
n
i=1
cos
π/2
i+1
converges to a strictly positive limit. In fact, in the exercises it is shown that the limit
is 2/π .
At this point we have two necessary and sufficient conditions for convergence
of infinite products. However, these results apply to particular infinite products (the
general term must be always above 1 or always below 1). We now turn to a condition
for convergence that applies to any infinite product but is only a sufficient condition.
A sufficient condition for convergence
Let u
i
be a sequence of real numbers in (−1, +∞). If the sequence
p
n
=
n
i=1
1 +|u
i
|
converges, then the sequence
q
n
=
n
i=1
(1 +u
i
)
converges to a strictly positive limit.
The result above is closely related to the corresponding result for series: absolute
convergence implies convergence.
We first need two simple lemmas.
126 4 Fifty Ways to Estimate the Number π
Lemma 4.1 For x in [0, 1), we have
ln(1 +x) ≤
x
1 −x
.
To prove Lemma 4.1, we start with the power series expansion
ln(1 +x) =
∞
n=1
(−1)
n+1
x
n
/n =x −x
2
/2 +x
3
/3 +···,
which holds for all x in [−1, 1). Observe now that for x ≥0 and n ≥1, we have
(−1)
n+1
x
n
/n ≤x
n
/n ≤x
n
.
Thus,
ln(1 +x) ≤
∞
n=1
x
n
=
x
1 −x
for all x ∈[0, 1).
This proves Lemma 4.1.
Lemma 4.2 For a ll y in [−1/4, 1/4], we have
ln(1 +y)
≤2|y|.
If y ≥0, we have, by Lemma 4.1,
ln(1 +y)
=ln(1 +y) ≤
y
1 −y
.
If y ≤1/4, then 1 −y ≥1 −1/4 and
1
1 −y
≤4/3 < 2.
Hence,
y
1 −y
≤4y/3 < 2y
and
ln(1 +y)
≤2y =2|y| for all y ∈[0, 1/4].
We now turn to y in [−1/4, 0). Since 1 +y<1, ln(1 +y) < 0. Therefore,
ln(1 +y)
=−ln(1 +y) =ln
1
1 +y
=ln
1 −
y
1 +y
.
Using that y<0, we get
y
1 +y
=
−|y|
1 −|y|
.
Hence,
ln(1 +y)
=ln
1 +
|y|
1 −|y|
.
4.3 Convergence of Infinite Products 127
Note that if |y|< 1/2, then
|y|
1 −|y|
< 1,
and we may apply Lemma 4.1 to
x =
|y|
1 −|y|
to get
ln(1 +y)
=ln
1 +
|y|
1 −|y|
=ln(1 +x) ≤
x
1 −x
=
|y|
1 −2|y|
.
But
1
1 −2|y|
≤2for|y|≤1/4.
Therefore,
ln(1 +y)
≤2|y|.
This completes the proof of Lemma 4.2.
We are now ready to show that absolute convergence implies convergence for
infinite products. Assume that
p
n
=
n
i=1
1 +|u
i
|
converges. Since |u
i
|≥0, we know that if p
n
converges, then
∞
i=1
|u
i
|
converges. In particular, |u
i
| converges to 0, and there exists N such that if i ≥ N,
then
|u
i
|< 1/4.
Hence, by Lemma 4.2,
ln(1 +u
i
)
< 2|u
i
| for i ≥N.
By the series comparison test the positive terms series
∞
i=1
ln(1 +u
i
)
converges. Hence, the series
∞
i=1
ln(1 +u
i
)
128 4 Fifty Ways to Estimate the Number π
converges absolutely and therefore converges. Let
S
n
=
n
i=1
ln(1 +u
i
).
The sequence S
n
converges to some limit S. By the continuity of the exp function,
exp(S
n
) converges to exp(S).But
exp(S
n
) =
n
i=1
(1 +u
i
) =q
n
.
This proves that q
n
converges to exp(S) > 0. We have proved that absolute conver-
gence implies convergence for infinite products.
Example 4.4 Does the infinite product
n
i=2
1 +
(−1)
n
n
2
converge?
Since
n
i=2
(−1)
n
n
2
converges, so does
n
i=2
1 +
(−1)
n
n
2
,
and therefore,
n
i=2
1 +
(−1)
n
n
2
converges absolutely, and so it converges.
Example 4.5 Does the infinite product
n
i=2
1 +
(−1)
n
n
converge?
In this case the series
n
i=2
(−1)
n
n
4.3 Convergence of Infinite Products 129
diverges, so
n
i=2
1 +
(−1)
n
n
diverges as well. This shows that
n
i=2
1 +
(−1)
n
n
does not converge absolutely. This does not prove that it diverges. It actually con-
verges, but we need another criterion to prove it. See the exercises.
Exercises
1. Consider the sequence
q
n
=
n
i=1
1 +
1
i
p
.
Find the values of p for which q
n
converges.
2. Find the limit of
n
i=2
1 −
1
i
.
3. Find the values of p for which
n
i=2
1 −
1
i
p
converges to a strictly positive limit.
4. Assume that |x|< 1.
(a) Show that
ln(1 +x) −x
<
x
2
2(1 −|x|)
.
(b) Use (a) to show that if |x|< 1/2, then
ln(1 +x) −x
<x
2
.
5. In this exercise we prove another necessary and sufficient condition for conver-
gence for infinite products. Consider
∞
n=1
(1 + a
n
) with all a
n
> −1 and such
that
∞
n=1
a
2
n
converges.
Under this condition, we show that
∞
n=1
(1 +a
n
) converges to a nonzero limit
if and only if
∞
n=1
a
n
converges.