Correct answer!
We use the .
For x ³ 2, the function
is continuous, positive and decreasing. () We find an antiderivative of f using
substitution u=3x-1-1 ()
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|
ó
õ
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1
3x-1-1
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dx = |
1
ln(3)
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ó
õ
|
|
du
u(u+1)
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|
|
Expanding into partial fractions we obtain ()
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1
ln(3)
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|
ó
õ
|
|
æ
ç
è
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|
1
u
|
- |
1
u+1
|
|
ö
÷
ø
|
du = |
1
ln(3)
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( ln(3x-1-1)-ln(3x-1)) +C |
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We now evaluate the improper integral:
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ó
õ
|
¥
2
|
|
1
3x-1-1
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dx = |
1
ln(3)
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|
é
ê
ë
|
ln( |
3x-1-1
3x-1
|
) |
ù
ú
û
|
¥
2
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=
|
lim
b® ¥
|
|
1
ln(3)
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é
ë
| ln(1-31-x) |
ù
û
|
b
2
|
=
|
-ln(2/3)
ln(3)
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|
|
Since this integral converges, by the our series converges
.
Note that this problem could also be solved using