Page 264 - Calculus Workbook For Dummies

248 Part IV: Integration and Infinite Series

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3
Q. Does ! 1 converge or diverge? Q. Does ! 1 converge or diverge?
2
n 2 n - n n 2 n lnn
=
=
A. It converges. A. It diverges.
1. Try the nth term test. Tip: If you can see that you’ll be able
1 to integrate the series expression,
No good: lim 2 = 0 you’re home free. So always ask yourself
n " 3 n - n
whether you can use the integral com-
2. Try the direct comparison test. parison test.
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3
! 1 resembles ! 1 2 , which you
2
n 2 n - n n 2 n 1. Ask yourself whether you know how to
=
=
know converges by the p-series rule. But integrate this expression.
the direct comparison test is no help Sure. It’s an easy u-substitution.
3
because each term of ! 1 is greater
2
n 2 n - n 2. Do the integration.
=
than your known convergent series.
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3
3. Try the limit comparison test with ! 1 2 . # 1 dx
=
Piece o’ cake. n 2 n 2 x lnx
c 1
It’s best to put your known, benchmark = lim # dx
c " 3 x lnx
series in the denominator. 2
1 u = lnx when x = 2 , u = ln2
2
lim n - n
n " 3 1 du = 1 dx when x = , c u = lnc
n 2 x
n 2
= lim ln c
2
n " 3 n - n lim # - / 1 2
= u du
= 1 _ By the horizontalasymptote rulei c " 3 ln 2
ln c
2 7
/ 1 2
= lim u A
Because the limit is finite and positive c " 3 ln 2
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3
and because ! 1 2 converges, ! 1 = lim 2 ` lnc - 2 ln2j
2
=
n 2 n n 2 n - n c " 3
=
also converges. = 3
Because this improper integral diverges,
so does the companion series.