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11.29
Definition (Alternating series.)
Series of the form
![$\displaystyle {\sum\{(-1)^na_n\}}$](img1523.gif)
or
![$\displaystyle {\sum\{(-1)^{n+1}a_n\}}$](img1524.gif)
where
![$a_j\geq 0$](img1525.gif)
for all
![$j$](img1526.gif)
are called
alternating series.
11.30
Theorem (Alternating series test.)
Let
be a decreasing sequence of positive numbers such that
. Then
is summable. Moreover,
and
for all
.
Proof: Let
. For all
,
and
Thus
is decreasing and
is increasing. Also, for all
,
so
is bounded below by
, and
so
is bounded above by
.
It follows that there exist real numbers
and
such that
Now
so
.
It follows from the next lemma that
; i.e.,
Since for all
we have
and since
Thus, in all cases,
; i.e.,
approximates
with an error of no more than
.
11.31
Lemma.
Let
be a real sequence and let
. Suppose
and
. Then
.
Proof: Let
be a precision function for
and let
be a precision
function for
. For all
, define
I claim
is a precision function for
, and hence
.
Let
.
- Case 1:
is even. Suppose
is even.
Say
where
. Then
- Case 2:
is odd.
Suppose
is odd. Say
where
.
Then
Hence, in all cases,
11.32
Remark.
The alternating series test has obvious generalizations for series such as
and we will use these generalizations.
11.33
Example.
If
![$0\leq t\leq 1$](img1566.gif)
, then
are decreasing positive null sequences, so
are summable; i.e.,
(These are the sequences we called
![$\{C_n(t)\}$](img1293.gif)
and
![$\{S_n(t)\}$](img1296.gif)
in example
10.3.)
Also,
,
with an error smaller than
. My calculator says
and
11.34
Entertainment.
Since
![$\displaystyle {\left\{ {{t^n}\over n}\right\}_{n\geq 1}}$](img1574.gif)
is a decreasing positive null
sequence for
![$0\leq t\leq 1$](img1566.gif)
, it follows that
![$\displaystyle {\sum\left\{{{(-1)^{n-1}t^n}\over
n}\right\}_{n\geq 1}}$](img1575.gif)
converges for
![$0\leq t\leq 1$](img1566.gif)
. We will now explicitly
calculate the limit of this series using a few ideas that are not justified by results
proved in this course. We know that for all
![$x\in\mbox{{\bf R}}\setminus \{-1\}$](img1576.gif)
, and all
![$n\in\mbox{{\bf N}}$](img9.gif)
,
Hence, for all
![$t>-1$](img1578.gif)
,
i.e.,
Thus
Hence
for all
![$t>-1$](img1578.gif)
.
If we can show that
is a null
sequence, it follows that
or in other words,
![\begin{displaymath}
\ln(1+t)=\sum_{j=1}^\infty{{(-1)^{j+1}t^j}\over j}.
\end{displaymath}](img1585.gif) |
(11.35) |
I claim
![$\displaystyle {\left\{\int_0^t{{x^n}\over {1+x}}dx\right\}}$](img1583.gif)
is a null sequence for
![$-1<t\leq 1$](img1586.gif)
and hence (
11.35) holds for
![$-1<t\leq 1$](img1586.gif)
. In particular,
First suppose
![$t\geq 0$](img1588.gif)
, then
![$\displaystyle {{1\over {1+x}}x^n\leq x^n}$](img1589.gif)
for
![$0\leq x\leq t$](img1590.gif)
, so
Since
![$\displaystyle { \left\{ {{t^{n+1}}\over {n+1}}\right\}}$](img1592.gif)
is a null sequence for
![$0\leq t\leq 1$](img1566.gif)
, it follows from the comparison test that
![$\displaystyle {\left\{\int_0^t{{x^n}\over {1+x}}dx\right\}}$](img1583.gif)
is a null sequence for
![$0\leq t\leq 1$](img1566.gif)
.
Now suppose
![$-1<t<0$](img1593.gif)
. Then
so
![$\displaystyle { {{\vert x\vert^n}\over {1+x}}\leq {{\vert x\vert^n}\over {1+t}}}$](img1595.gif)
and
If
![$-1<t<0$](img1593.gif)
, then
![$\displaystyle { \left\{{1\over {1+t}} \cdot{{\vert t\vert^{n+1}}\over {n+1}}\right\}}$](img1597.gif)
is a
null sequence, so
![$\displaystyle {\left\{\int_0^t{{x^n}\over {1+x}}dx\right\}}$](img1583.gif)
is a null
sequence.
11.36
Entertainment.
By starting with the formula
for all
![$x\in\mbox{{\bf R}}$](img699.gif)
and using the ideas from the last example, show that
![\begin{displaymath}
\sum_{j=0}^\infty {{(-1)^jx^{2j+1}}\over {(2j+1)}}=\arctan(x)\mbox{ for all }x\in[-1,1].
\end{displaymath}](img1599.gif) |
(11.37) |
Conclude that
Next: 11.4 Absolute Convergence
Up: 11. Infinite Series
Previous: 11.2 Convergence Tests
  Index