Download Telescoping and Geometric Series Lab

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Section:
Telescoping and Geometric Series Lab
Introduction: In this lab we will evaluate infinite geometric and telescoping series. Non-trivial geometric series are series
∞
∑
that can be written in the form
ark where a and r are non-zero real numbers. Recall that if |r| < 1 the geometric
k=0
a
series sums to 1−r
, and if |r| ≥ 1 then the series diverges. Very loosely speaking, telescoping series are series for which
you can define the nth partial sum of the series, denoted Sn , in such a way that lim Sn can be evaluated exactly, and
n→∞
whose partial sums eventually have only a fixed number of terms after cancellation. The following is an example of a
series composed of both types.
Ex: Find the sum of the series
∞
∑
[
k=2
Solution:
Note: In order to use the equality
∞
∑
k=2
[
7
1
− k−1
2
k − 1 10
]
1
k2 −1
− 10k−1
7
=
∞
∑
k=2
1
k2 −1
−
]
∞
∑
k=2
7
,
10k−1
we must first show that each series on
the right side of the equality independently converges. The first term on the right hand side is a telescoping series
and the second term is a geometric series. We may simply subtract their sums to arrive at our answer.
(a) Using the technique of partial fractions we find that
1
1
1
1
=
=
−
k2 − 1
(k + 1)(k − 1)
2k − 2 2k + 2
=
1[ 1
1 ]
−
2 k−1 k+1
and thus we have that
∞
∑
k=2
∑
1 ∑[ 1
1 ]
1
1
=
=
−
2
k −1
(k + 1)(k − 1)
2
k−1 k+1
∞
∞
k=2
k=2
using the following telescoping technique for n > 2 we can try to find a pattern for Sn :
1
S2 =
2
(
1
1−
3
)
[(
) (
)]
(
)
1
1 1
1 3 1
1
1
1−
+
−
=
− −
S3 = S2 + a3 =
2
3
2 4
2 2 3 4
[(
) (
)]
(
)
3 1 1
1 3 1
1
1 1
1
S4 = S3 + a4 =
− −
+
=
− −
−
2
2 3 4
3 5
2 2 4 5
[(
) (
)]
(
)
3 1 1
1 3 1 1
1
1 1
− −
+
=
− −
S5 =
−
2
2 4 5
4 6
2 2 5 6
1∑
1(3
1
1 )
1
= Sn =
− −
2
(k + 1)(k − 1)
2 2 n n+1
n
⇒
k=2
Notice that, due to cancellations, the partial sum has only a small number of terms. Now we may easily take
1
a limit as follows:
] 3
1(3
1
1 ) 1[3
− −
=
−0−0 =
n→∞ 2 2
n n+1
2 2
4
lim Sn = lim
n→∞
⇒
∞
∑
k=2
1
3
= lim Sn =
n→∞
−1
4
k2
(b) Moving on to the second series we wish to rearrange the terms into a more familiar geometric series form as
follows:
( )k
∞
∞
∞
∞
∑
∑
∑
( 1 )k+1 ∑
7
7
7
1
=
=
7
=
10k−1
10k
10
10 10
k=2
This series is convergent since
k=1
1
10
< 1 and a =
k=0
7
10 .
k=0
This implies that the series converges as follows:
∞
7
∑
7 ( 1 )k
7
= 10 1 =
10 10
9
1
−
10
k=0
(c) Now since both series converge we can answer the original problem to get:
[
]
∞
∑
1
7
3 7
−
= −
k 2 − 1 10k−1
4 9
k=2
= −
1
36
∞
∑
1
,
(k + a)(k + b)
k=1
where a and b are real numbers. We will endeavor to discover for which values of a and b we may evaluate the series
using telescoping and for which values we may use integration. For the purposes of this lab we will assume that
a ̸= b. In all that follows then we can assume, without loss of generality, that a > b.
• Directions: In this lab we will be exploring a family of infinite series of the form S(a, b) =
1. Show that
∞
1 ∑
S(a, b) =
a−b
k=1
2
(
1
1
−
k+b k+a
)
2. (a) Consider the series S(1, 0). Write out the first 3 terms, and then the nth term of the series Sn .
(b) Find the sum of the series using a telescoping technique.
3. (a) Write out the first couple terms using partial fraction decomposition, and then the nth term of the series S(2, 0).
(b) Find the sum of the series using a telescoping technique.
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4. Suppose that a − b = m, such that m ∈ N.
(a) Write out the first couple terms of the series S(a, b) using partial fraction decomposition. Use the fact that
b = a − m.
(b) Show that all but m terms of the series cancel and that the series reduces to a finite sum with m terms.
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(c) Show that when a and b are non-negative integers with a − b = m, we have
∞
S(a, a − m) =
1 ∑
m
k=1
(d) Use part (c) to confirm that S(4, 1) =
(
1
1
−
k+a−m k+a
13
.
36
5
)
1 ∑
1
m
k+a−m
m
=
k=1
Challenge:
i. Assume a and b are non-negative distinct numbers. Make a convincing argument for why S(a, b) must
1
1
converge. Hint: Use the face that
< 2 for every k ∈ N.
(k + a)(k + b)
k
ii. Assume a = b. Make a conjecture for the convergence of S(a, b) and justify your claim.
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