8.2 Sequences
I. Find the limit of the following sequences, or determine that the limit does not
exist:
3n3 − 1
}
2n3 + 1
Divide numerator and denominator by n3 , we get:
1. {
3n3 − 1
n→∞ 2n3 + 1
=
lim
=
=
lim
n→∞
3−
2+
1
n3
1
n3
limn→∞ (3 −
limn→∞ (2 +
3
2
1
)
n3
1
)
n3
tan−1 n
}
n
π
As n → ∞, tan−1 n → , so
2
2. {
tan−1 n
n→∞
n
limn→∞ (tan−1 n)
limn→∞ n
= 0
lim
=
1 n
)2}
2n
n
1
1 n
) 2 = e 2 ln(1+ 2n ) ,
Since (1 +
2n
Take the logarithm of the expression and use L’Hospital rule:
3. {(1 +
1
n
ln(1 +
) =
n→∞ 2
2n
lim
=
lim
n→∞
lim
ln(1 +
1
2n )
2
n
1
1+(1/2n)
·
−1
2n2
−2/n2
1
= lim
n→∞ 4(1 + (1/2n))
1
=
4
1
Thus, the original limit is e 4
1
n→∞
6
4. {n sin( )}.
n
Since the expression is of the form ∞ · 0, we use the L’Hospital rule.
6
lim n sin( ) =
n→∞
n
=
lim
sin( n6 )
1
n
−6 cos(6/n)
n2
lim
n→∞ (−1/n2 )
n→∞
6
lim 6 cos( )
n→∞
n
= 6
=
5.
n
n+1
−n+1
n+1
n is even
n is odd.
2n
, which converges to 1.
2n + 1
−n
The odd terms form a sequence b2n+1 =
, which converges to −1.
n+1
Thus the sequence as a whole does not converge.
The even terms form a sequence b2n =
II. Determine whether the following sequence converges or diverges and describe
whether it does so monotonically or by oscillation. Give the limit when the
sequence converges.:
1. {(−0.7)n }
Since | − 0.7| < 1, the sequence converges to 0;
Since −0.7 < 0, it converges monotonically.
2. {(1.00001)n }
Since 1.00001 > 1, the sequence diverges;
Since 1.00001 > 0, the divergence is monotone.
2