CK-12-Calculus

http://www.ck12.org Chapter 8. Infinite Series

Sn=

n

k∑= 1

( 2

k−

2

k+ 1

)

=

( 2

1 −

2

1 + 1

)

+

( 2

2 −

2

2 + 1

)

+

( 2

3 −

2

3 + 1

)

+...+

( 2

n−

2

n+ 1

)

=

( 2

1 −

2

2

)

+

( 2

2 −

2

3

)

+

( 2

3 −

2

4

)

+...+

( 2

n−

2

n+ 1

)

We can simplifysnfurther. Notice that the first parentheses has−^22 while the second parentheses has^22. These will
add up to 0 and cancel out. Likewise, the−^23 and^23
will cancel out. Continue in this way to cancel opposite terms. This sum is atelescoping sum, which is a sum of
terms that cancel each other out so that the sum will fold neatly like a folding telescope. Thus, we can write the
partial sum as

sn=^21 −n+^21 = 2 −n+^21.

Then limn→+∞sn=limn→+∞( 2 −n+^21 )=2 and∑+k=∞ 1 (^2 k−k+^21 )=2.
Other Divergent Series (nth-Term Test)
Determining convergence by using the limit of the sequence of partial sums is not always feasible or practical. For
other series, it is more useful to apply tests to determine if an infinite series converges or diverges. Here are two
theorems that help us determine convergence or divergence.
Theorem (Thenth-Term Test)
If the infinite series∑∞k= 1 ukconverges, then limk→+∞uk= 0
Theorem
If limk→+∞uk 6 =0 or limk→+∞ukdoes not exist, then the infinite series∑∞k= 1 ukdiverges.
The first theorem tells us that if an infinite series converges, then the limit of the sequence of terms is 0. The converse
is not true: If the limit of the sequence of terms is 0, then the series converges. So, we cannot use this theorem as a
test of convergence.
The second theorem tells us that if limit of the sequence of terms is not zero, then the infinites series diverges. This
gives us the first test of divergence: thenth-Term TestorDivergence Test. Note that if the test is applied and the
limit of the sequence of terms is 0, we cannot conclude anything and must use another test.
Example 9

Determine if∑∞k= (^1) k+k 5 converges or diverges.
Solution
We can use thenth-Term Test to determine if the series diverges. Then we do not have to check for convergence.
k→lim+∞k+k 5 =k→lim+∞
kk
k+k 5 =klim→+∞

1

1 +^5 k=

limk→+∞ 1

limk→+∞ 1 +^5 k=^1

Because limk→+∞ k+k 56 =0, the series∑∞k= (^1) k+k 5 diverges.
Example 10