Begin2.DVI

Example 12-2. (Sum of series)

Show that

1

2 · 5

+^1

8 · 11

+^1

14 · 17

+^1

20 · 23

+··· =^1

9

(π

3

+ ln2)

Solution

Let Sdenote the sum of the series and use partial fractions to write

1

n·(n+ 3)

=A

n

+ B

n+ 3

for n= 2, 8 , 14 , 20 ,... to show that

S=^13

[

1

2 −

1

5 +

1

8 −

1

11 +

1

14 −

1

17 +

1

20 −

1

23 +···

]

Note that the sum Sis a special case of the Taylor’s series

S(x) =^1

3

[

x^2

2

−x

5

5

+x

8

8

−x

11

11

+x

14

14

−x

17

17

+x

20

20

−x

23

23

+···

]

with S=S(1) the desired sum. The derivative of S(x)produces

dS

dx

=^1

3

[

x−x^4 +x^7 −x^10 +x^13 −x^16 +x^19 −x^22 +···

]

The derivative series is recognized as a geometric series with sum x

x^3 + 1

so that one

can write

dS

dx

=^1

3

x

x^3 + 1

The desired series sum can now be expressed in terms of an integral

S=S(1) =

1

3

∫ 1

0

x

x^3 + 1 dx

As an exercise, use partial fractions and show

S=S(1) =^1

3

∫ 1

0

x

x^3 + 1

dx =^1

3

[

√^1

3

tan−^1

(

2 x√− 1

3

)

−^1

3

ln(x+ 1) +^1

6

ln(1 −x+x^2 )

] 1

0

which simplifies to

S=S(1) =^1

9

(

√π

3

−ln 2

)

Refraction through a prism

Refraction through a prism is often encountered in physics courses and the calcu-

lus needed to analyze the physical problem is messy. Let’s investigate this problem.