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The Har= monic Series and the Integral Test
------=_NextPart_01C9C1B6.6161D560 Content-Location: file:///C:/1A7BA2D4/25TheHarmonicSeriesandtheIntegralTest_files/slide0002.htm Content-Transfer-Encoding: quoted-printable Content-Type: text/html; charset="us-ascii" PowerPoint Presentation
The Har= monic Series and the Integral Test
If = we add infinitely many terms and obtain a finite sum, it must be the case that the te= rms get smaller and smaller and goes to zero. 
------=_NextPart_01C9C1B6.6161D560 Content-Location: file:///C:/1A7BA2D4/25TheHarmonicSeriesandtheIntegralTest_files/slide0008.htm Content-Transfer-Encoding: quoted-printable Content-Type: text/html; charset="us-ascii" PowerPoint Presentation
Theorem:
The Har= monic Series and the Integral Test
If = we add infinitely many terms and obtain a finite sum, it must be the case that the te= rms get smaller and smaller and goes to zero. 
If        =3D a1 + a2 + a3 + … =3D L is a
Σ
i=3D1
∞
ai
convergent series, then lim an =3D 0.
nà∞
------=_NextPart_01C9C1B6.6161D560 Content-Location: file:///C:/1A7BA2D4/25TheHarmonicSeriesandtheIntegralTest_files/slide0009.htm Content-Transfer-Encoding: quoted-printable Content-Type: text/html; charset="us-ascii" PowerPoint Presentation
Proof: Let       =3D= a1 + a2 .. =3D L be a convergent series. <= /span>
Σ
i=3D1
∞
ai
Theorem:
The Har= monic Series and the Integral Test
If = we add infinitely many terms and obtain a finite sum, it must be the case that the te= rms get smaller and smaller and goes to zero. 
If        =3D a1 + a2 + a3 + … =3D L is a
Σ
i=3D1
∞
ai
convergent series, then lim an =3D 0.
nà∞
------=_NextPart_01C9C1B6.6161D560 Content-Location: file:///C:/1A7BA2D4/25TheHarmonicSeriesandtheIntegralTest_files/slide0010.htm Content-Transfer-Encoding: quoted-printable Content-Type: text/html; charset="us-ascii" PowerPoint Presentation
Proof: Let       =3D= a1 + a2 .. =3D L be a convergent series. <= /span>
Σ
i=3D1
∞
ai
Theorem:
This means the for the sequence of partial sums,
<= /div>
lim sn =3D lim (a1 + a2 + … + an) =3D= L converges. 
The Har= monic Series and the Integral Test
If = we add infinitely many terms and obtain a finite sum, it must be the case that the te= rms get smaller and smaller and goes to zero. 
nà∞
If        =3D a1 + a2 + a3 + … =3D L is a
Σ
i=3D1
∞
ai
convergent series, then lim an =3D 0.
nà∞
------=_NextPart_01C9C1B6.6161D560 Content-Location: file:///C:/1A7BA2D4/25TheHarmonicSeriesandtheIntegralTest_files/slide0011.htm Content-Transfer-Encoding: quoted-printable Content-Type: text/html; charset="us-ascii" PowerPoint Presentation
Proof: Let       =3D= a1 + a2 .. =3D L be a convergent series. <= /span>
Σ
i=3D1
∞
ai
Theorem:
This means the for the sequence of partial sums,
<= /div>
lim sn =3D lim (a1 + a2 + … + an) =3D= L converges. 
The Har= monic Series and the Integral Test
If = we add infinitely many terms and obtain a finite sum, it must be the case that the te= rms get smaller and smaller and goes to zero. 
nà∞
lim sn-1 =3D lim (a1 + a2 +..+ an-1) <= /span>
nà∞
nà∞
If        =3D a1 + a2 + a3 + … =3D L is a
Σ
i=3D1
∞
ai
convergent series, then lim an =3D 0.
nà∞
On the other hand, 
------=_NextPart_01C9C1B6.6161D560 Content-Location: file:///C:/1A7BA2D4/25TheHarmonicSeriesandtheIntegralTest_files/slide0012.htm Content-Transfer-Encoding: quoted-printable Content-Type: text/html; charset="us-ascii" PowerPoint Presentation
Proof: Let       =3D= a1 + a2 .. =3D L be a convergent series. <= /span>
Σ
i=3D1
∞
ai
Theorem:
This means the for the sequence of partial sums,
<= /div>
lim sn =3D lim (a1 + a2 + … + an) =3D= L converges. 
The Har= monic Series and the Integral Test
If = we add infinitely many terms and obtain a finite sum, it must be the case that the te= rms get smaller and smaller and goes to zero. 
nà∞
lim sn-1 =3D lim (a1 + a2 +..+ an-1) = =3D L.
nà∞
nà∞
If        =3D a1 + a2 + a3 + … =3D L is a
Σ
i=3D1
∞
ai
convergent series, then lim an =3D 0.
nà∞
On the other hand, 
------=_NextPart_01C9C1B6.6161D560 Content-Location: file:///C:/1A7BA2D4/25TheHarmonicSeriesandtheIntegralTest_files/slide0013.htm Content-Transfer-Encoding: quoted-printable Content-Type: text/html; charset="us-ascii" PowerPoint Presentation
Proof: Let       =3D= a1 + a2 .. =3D L be a convergent series. <= /span>
Σ
i=3D1
∞
ai
Theorem:
This means the for the sequence of partial sums,
<= /div>
lim sn =3D lim (a1 + a2 + … + an) =3D= L converges. 
The Har= monic Series and the Integral Test
If = we add infinitely many terms and obtain a finite sum, it must be the case that the te= rms get smaller and smaller and goes to zero. 
nà∞
lim sn-1 =3D lim (a1 + a2 +..+ an-1) = =3D L.
nà∞
nà∞
Hence lim sn – sn-1 
nà∞
If        =3D a1 + a2 + a3 + … =3D L is a
Σ
i=3D1
∞
ai
convergent series, then lim an =3D 0.
nà∞
On the other hand, 
------=_NextPart_01C9C1B6.6161D560 Content-Location: file:///C:/1A7BA2D4/25TheHarmonicSeriesandtheIntegralTest_files/slide0014.htm Content-Transfer-Encoding: quoted-printable Content-Type: text/html; charset="us-ascii" PowerPoint Presentation
Proof: Let       =3D= a1 + a2 .. =3D L be a convergent series. <= /span>
Σ
i=3D1
∞
ai
Theorem:
This means the for the sequence of partial sums,
<= /div>
lim sn =3D lim (a1 + a2 + … + an) =3D= L converges. 
The Har= monic Series and the Integral Test
If = we add infinitely many terms and obtain a finite sum, it must be the case that the te= rms get smaller and smaller and goes to zero. 
nà∞
lim sn-1 =3D lim (a1 + a2 +..+ an-1) = =3D L.
nà∞
nà∞
Hence lim sn – sn-1 =3D lim an 
nà∞
nà∞
If        =3D a1 + a2 + a3 + … =3D L is a
Σ
i=3D1
∞
ai
convergent series, then lim an =3D 0.
nà∞
On the other hand, 
------=_NextPart_01C9C1B6.6161D560 Content-Location: file:///C:/1A7BA2D4/25TheHarmonicSeriesandtheIntegralTest_files/slide0015.htm Content-Transfer-Encoding: quoted-printable Content-Type: text/html; charset="us-ascii" PowerPoint Presentation
Proof: Let       =3D= a1 + a2 .. =3D L be a convergent series. <= /span>
Σ
i=3D1
∞
ai
Theorem:
This means the for the sequence of partial sums,
<= /div>
lim sn =3D lim (a1 + a2 + … + an) =3D= L converges. 
The Har= monic Series and the Integral Test
If = we add infinitely many terms and obtain a finite sum, it must be the case that the te= rms get smaller and smaller and goes to zero. 
nà∞
lim sn-1 =3D lim (a1 + a2 +..+ an-1) = =3D L.
nà∞
nà∞
Hence lim sn – sn-1 =3D lim an =3D L – L =3D 0.
nà∞
nà∞
If        =3D a1 + a2 + a3 + … =3D L is a
Σ
i=3D1
∞
ai
convergent series, then lim an =3D 0.
nà∞
On the other hand, 
------=_NextPart_01C9C1B6.6161D560 Content-Location: file:///C:/1A7BA2D4/25TheHarmonicSeriesandtheIntegralTest_files/slide0003.htm Content-Transfer-Encoding: quoted-printable Content-Type: text/html; charset="us-ascii" PowerPoint Presentation
The Har= monic Series and the Integral Test
How= ever the fact that lim anà 0 does not guarantee that their sums converge.
------=_NextPart_01C9C1B6.6161D560 Content-Location: file:///C:/1A7BA2D4/25TheHarmonicSeriesandtheIntegralTest_files/slide0016.htm Content-Transfer-Encoding: quoted-printable Content-Type: text/html; charset="us-ascii" PowerPoint Presentation
Example:
The sequence
The Har= monic Series and the Integral Test
How= ever the fact that lim anà 0 does not guarantee that their sums converge.
1
2
,
1
2
,
1
3
,
1
3
,
1
3
,
1
4
,
1
4
,
1
4
,
1
4
,
à 0,
1
5
, ..<= /span>
------=_NextPart_01C9C1B6.6161D560 Content-Location: file:///C:/1A7BA2D4/25TheHarmonicSeriesandtheIntegralTest_files/slide0017.htm Content-Transfer-Encoding: quoted-printable Content-Type: text/html; charset="us-ascii" PowerPoint Presentation
Example:
The sequence
The Har= monic Series and the Integral Test
How= ever the fact that lim anà 0 does not guarantee that their sums converge.
1
2
,
1
2
,
1
3
,
1
3
,
1
3
,
1
4
,
1
4
,
1
4
,
1
4
,
à 0,
1
5
, ..<= /span>
but their sum
1
2
+
1
2
1
3
1
3
1
3
1
4
1
4
1
4
1
4
1
5
..
+
+
+
+
+
+
+
+
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Content-Type: image/gif R0lGODlhDwACAHcAMSH+GlNvZnR3YXJlOiBNaWNyb3NvZnQgT2ZmaWNlACH5BAEAAAAALAAAAAAO AAEAgAAAAAAAAAIDjI9YADt= ------=_NextPart_01C9C1B6.6161D560 Content-Location: file:///C:/1A7BA2D4/25TheHarmonicSeriesandtheIntegralTest_files/slide0017_image004.gif Content-Transfer-Encoding: base64 Content-Type: image/gif R0lGODlhDwACAHcAMSH+GlNvZnR3YXJlOiBNaWNyb3NvZnQgT2ZmaWNlACH5BAEAAAAALAAAAAAP AAEAgAAAAAAAAAIDjI9ZADt= ------=_NextPart_01C9C1B6.6161D560 Content-Location: file:///C:/1A7BA2D4/25TheHarmonicSeriesandtheIntegralTest_files/slide0017_image005.gif Content-Transfer-Encoding: base64 Content-Type: image/gif R0lGODlhDwACAHcAMSH+GlNvZnR3YXJlOiBNaWNyb3NvZnQgT2ZmaWNlACH5BAEAAAAALAEAAAAO AAEAgAAAAAAAAAIDjI9YADt= ------=_NextPart_01C9C1B6.6161D560 Content-Location: file:///C:/1A7BA2D4/25TheHarmonicSeriesandtheIntegralTest_files/slide0017_image006.gif Content-Transfer-Encoding: base64 Content-Type: image/gif R0lGODlhEAACAHcAMSH+GlNvZnR3YXJlOiBNaWNyb3NvZnQgT2ZmaWNlACH5BAEAAAAALAEAAAAO 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Example:
The sequence
The Har= monic Series and the Integral Test
How= ever the fact that lim anà 0 does not guarantee that their sums converge.
1
2
,
1
2
,
1
3
,
1
3
,
1
3
,
1
4
,
1
4
,
1
4
,
1
4
,
à 0,
1
5
, ..<= /span>
but their sum
1
2
+
1
2
1
3
1
3
1
3
1
4
1
4
1
4
1
4
1
5
..
+
+
+
+
+
+
+
+
=3D <= /span>∞
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AAEAgAAAAAAAAAIDjI9YADt= ------=_NextPart_01C9C1B6.6161D560 Content-Location: file:///C:/1A7BA2D4/25TheHarmonicSeriesandtheIntegralTest_files/slide0018_image040.gif Content-Transfer-Encoding: base64 Content-Type: image/gif R0lGODlhEAACAHcAMSH+GlNvZnR3YXJlOiBNaWNyb3NvZnQgT2ZmaWNlACH5BAEAAAAALAEAAAAO AAEAgAAAAAAAAAIDjI9YADt= ------=_NextPart_01C9C1B6.6161D560 Content-Location: file:///C:/1A7BA2D4/25TheHarmonicSeriesandtheIntegralTest_files/slide0019.htm Content-Transfer-Encoding: quoted-printable Content-Type: text/html; charset="us-ascii" PowerPoint Presentation
Example:
The sequence
The Har= monic Series and the Integral Test
How= ever the fact that lim anà 0 does not guarantee that their sums converge.
1
2
,
1
2
,
1
3
,
1
3
,
1
3
,
1
4
,
1
4
,
1
4
,
1
4
,
à 0,
1
5
, ..<= /span>
but their sum
1
2
+
1
2
1
3
1
3
1
3
1
4
1
4
1
4
1
4
1
5
..
+
+
+
+
+
+
+
+
=3D <= /span>∞
The most important sequence that goes to 0 but sum
<= /div>
to    is the harmonic sequence:=
∞
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Content-Transfer-Encoding: base64 Content-Type: image/gif R0lGODlhEAACAHcAMSH+GlNvZnR3YXJlOiBNaWNyb3NvZnQgT2ZmaWNlACH5BAEAAAAALAEAAAAO AAEAgAAAAAAAAAIDjI9YADt= ------=_NextPart_01C9C1B6.6161D560 Content-Location: file:///C:/1A7BA2D4/25TheHarmonicSeriesandtheIntegralTest_files/slide0019_image054.gif Content-Transfer-Encoding: base64 Content-Type: image/gif R0lGODlhEAACAHcAMSH+GlNvZnR3YXJlOiBNaWNyb3NvZnQgT2ZmaWNlACH5BAEAAAAALAEAAAAO AAEAgAAAAAAAAAIDjI9YADt= ------=_NextPart_01C9C1B6.6161D560 Content-Location: file:///C:/1A7BA2D4/25TheHarmonicSeriesandtheIntegralTest_files/slide0019_image055.gif Content-Transfer-Encoding: base64 Content-Type: image/gif R0lGODlhDwACAHcAMSH+GlNvZnR3YXJlOiBNaWNyb3NvZnQgT2ZmaWNlACH5BAEAAAAALAEAAAAN AAEAgAAAAAAAAAIDjI9XADt= ------=_NextPart_01C9C1B6.6161D560 Content-Location: file:///C:/1A7BA2D4/25TheHarmonicSeriesandtheIntegralTest_files/slide0019_image056.gif Content-Transfer-Encoding: base64 Content-Type: image/gif R0lGODlhDwACAHcAMSH+GlNvZnR3YXJlOiBNaWNyb3NvZnQgT2ZmaWNlACH5BAEAAAAALAEAAAAN AAEAgAAAAAAAAAIDjI9XADt= ------=_NextPart_01C9C1B6.6161D560 Content-Location: file:///C:/1A7BA2D4/25TheHarmonicSeriesandtheIntegralTest_files/slide0019_image057.gif Content-Transfer-Encoding: base64 Content-Type: image/gif R0lGODlhDwACAHcAMSH+GlNvZnR3YXJlOiBNaWNyb3NvZnQgT2ZmaWNlACH5BAEAAAAALAEAAAAN AAEAgAAAAAAAAAIDjI9XADt= ------=_NextPart_01C9C1B6.6161D560 Content-Location: file:///C:/1A7BA2D4/25TheHarmonicSeriesandtheIntegralTest_files/slide0019_image058.gif Content-Transfer-Encoding: base64 Content-Type: image/gif R0lGODlhDwACAHcAMSH+GlNvZnR3YXJlOiBNaWNyb3NvZnQgT2ZmaWNlACH5BAEAAAAALAEAAAAO AAEAgAAAAAAAAAIDjI9YADt= ------=_NextPart_01C9C1B6.6161D560 Content-Location: file:///C:/1A7BA2D4/25TheHarmonicSeriesandtheIntegralTest_files/slide0019_image059.gif Content-Transfer-Encoding: base64 Content-Type: image/gif R0lGODlhEAACAHcAMSH+GlNvZnR3YXJlOiBNaWNyb3NvZnQgT2ZmaWNlACH5BAEAAAAALAEAAAAO AAEAgAAAAAAAAAIDjI9YADt= ------=_NextPart_01C9C1B6.6161D560 Content-Location: file:///C:/1A7BA2D4/25TheHarmonicSeriesandtheIntegralTest_files/slide0019_image060.gif Content-Transfer-Encoding: base64 Content-Type: image/gif R0lGODlhEAACAHcAMSH+GlNvZnR3YXJlOiBNaWNyb3NvZnQgT2ZmaWNlACH5BAEAAAAALAEAAAAO AAEAgAAAAAAAAAIDjI9YADt= ------=_NextPart_01C9C1B6.6161D560 Content-Location: file:///C:/1A7BA2D4/25TheHarmonicSeriesandtheIntegralTest_files/slide0020.htm Content-Transfer-Encoding: quoted-printable Content-Type: text/html; charset="us-ascii" PowerPoint Presentation
Example:
The sequence
The Har= monic Series and the Integral Test
How= ever the fact that lim anà 0 does not guarantee that their sums converge.
1
2
,
1
2
,
1
3
,
1
3
,
1
3
,
1
4
,
1
4
,
1
4
,
1
4
,
à 0,
1
5
, ..<= /span>
but their sum
1
2
+
1
2
1
3
1
3
1
3
1
4
1
4
1
4
1
4
1
5
..
+
+
+
+
+
+
+
+
=3D <= /span>∞
The most important sequence that goes to 0 but sum
<= /div>
to    is the harmonic sequence:= {an} = =3D
1
2
,
1
3
,
1
4
,
1
5
, ..<= /span>
1,
{
}
∞
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