Does it converge or diverge? If it converges, find its value (if possible).
1.
∞
X
1
√
n− n
n=2
P1
The terms of the sum go to zero. It looks similar to
n , which diverges.
We also note that the terms of the sum are positive. We compare them:
1√
n− n
lim
1
n→∞
n
2.
n
1
√ = lim
=1
n − n n→∞ 1 − √1n
P
The series diverges by the limit comparison test, with (1/n).
n
o
= lim
n→∞
n√
1+ n
In this case, we simply take the limit:
n
√ = lim
n→∞ 1 +
n n→∞
lim
√
√1
n
n
=∞
+1
The sequence diverges.
3.
∞
X
n2 + 1
n3 − 1
n=2
2
The terms of the sum go to zero, since there is an
Pn1 in the numerator,
3
and n in the denominator. In fact, it looks like
n , so we compare it
to that:
n2 −1
n3 − n
3
=1
lim n 1−1 = lim 3
n→∞ n − 1
n→∞
n
P1
Therefore, the series diverges by the limit comparison test, with
n.
√
∞
X
5−2 n
4.
n3
n=1
We can temporarily break this apart to see if the pieces converge:
√
∞
∞
∞ √
X
X
5−2 n X 5
n
=
−2
3
3
3
n
n
n
n=1
n=1
n=1
Both of these are p−series, the first with p = 3, the second with p = 52 ,
therefore they converge separately, and so the sum also converges.
5.
∞
X
(−6)n−1 51−n
n=1
First, let’s rewrite the terms of the sum:
n−1 1−n
(−6)
5
(−6)n−1
=
=
5n−1
1
−6
5
n−1 6.
so that this is a geometric series with r =
diverges.
n
o
−6
5 .
Since |r| > 1, this series
n!
(n+2)!
We first simplify:
n!
1
=
(n + 2)!
(n + 1)(n + 2)
so the limit as n → ∞ is 0.
∞
X
n
7.
ln
n+1
n=1
A little tricky... First, note that this can be written as ln(n) − ln(n+1 ).
Now, let’s write out the nth partial sum:
Sn = ln(1) − ln(2) + ln(2) − ln(3) + ln(3) − ln(4) + . . . + ln(n) − ln(n + 1)
with cancellations,
Sn = 0 − ln(n + 1)
Now, the limit of Sn as n → ∞ is −∞, so the sum diverges.
8.
∞
X
3n + 2n
6n
n=2
A sum of geometric series:
∞
∞ n
∞ n
X
X
X
3n + 2n
1
1
(1/2)2
(1/3)2
2
=
+
=
+
=
n
6
2
3
1
−
(1/2)
1
−
(1/3)
3
n=2
n=2
n=2
9.
sin
nπ
2
Write out the first few terms of the sequence:
1, 0, −1, 0, 1, 0, −1, . . .
so the sequence diverges.
10.
∞
X
1
n(n
+
1)(n
+ 2)
n=1
First, we see the terms go to zero like
1
n3 .
n3
=1
n→∞ n(n + 1)(n + 2)
lim
so the series converges by the limit comparison test.
2 11.
∞
X
sin2 (n)
√
n n
n=1
First, do the terms go to zero? The maximum value of the sine function
is 1, and all terms of the sum are positive, so:
sin2 (n)
1
≤ 3/2
3/2
n
n
so the terms do goP
to zero. Actually, we’ve also done a direct comparison
∞
1
with the p−series n=1 n3/2
, which converges.
12.
∞
X
n
(n
+
1)2n
n=1
It looks like the terms are going to zero like 21n , so let’s compare it to
P
(1/2)n , which is a convergent geometric series.
n
1
1
· n ≤ n
n+1 2
2
So the series converges by a direct comparison.
Evaluate, if possible.
Z
x3
1.
dx
3
x +1
Do long division first!
1
x3
=1− 3
3
x +1
x +1
Can we factor x3 + 1? We see x = −1 gives 0, so x + 1 can be factored
out. Using long division,
x3 + 1
= x2 − x + 1
x+1
so that x3 + 1 = (x + 1)(x2 − x + 1) (NOTE: On the exam, you will be
able to factor the polynomial easier than this!)
By Partial Fractions,
x3
1
1
1
1
2−x
=1−
=1+ ·
+ · 2
3
2
x +1
(x + 1)(x − x + 1)
3 x+1 3 x −x+1
Now you have to complete the square to finish things off, and after some
long algebra,
(2x − 1)
1
1
1
2
−1
√
− ln(x + 1)
x + ln(x − x + 1) − √ tan
6
3
3
3
NOTE: This was a complicated exercise! If you made it through this one,
you could probably stop now- You’re ready! There won’t be anything this
complex on the exam...
3 Z
2.
0
1
1
dx
2 − 3x
There is a vertical asymptote at x = 23 , so we need to split the integral
there:
Z T
Z 1
1
1
1
dx = lim −
dx + lim +
dx
T →2/3
T →2/3 2 − 3x
0 2 − 3x
0 2 − 3x
Integrate by taking u = 2 − 3x, and we get that the antiderivative is
− 31 ln |2 − 3x|. Now, take the limits- we’ll do one here:
1
1
1 1
1
1 1
lim − − ln
+ · = lim − + ln (2 − 3T ) + ·
3
2 − 3T
3 2 T →2/3
3
3 2
T →2/3
which diverges, since 0 is a vertical asymptote for ln(x).
Z
1
3.
dx
x4 − x2
Factor and use partial fractions:
Z
1
1 1
1
dx = + ln(x − 1) − ln(x + 1)
x2 (x − 1)
x 2
2
Z
4.
∞
1
dx
1
+
ex
1
Use u = 1 + ex , so du = ex dx, so that du = (u − 1)dx.
Substitution gives:
Z
1
du =
u(u − 1)
Z
−1
1
+
du
u
u−1
Antidifferentiate, and we get:
− ln (1 + ex ) + ln(ex ) = ln
ex
1 + ex
Take the appropriate limit to get an answer of ln(2)
Z ∞
dx
5.
dx
2 (x + 2)
(x
+
1)
0
Use partial fractions:
Z
Z
dx
1
1
1
dx =
+
−
dx
2
2
(x + 1) (x + 2)
(x + 1)
x+2 x−1
Antidifferentiate to get:
−
1
1
x+2
+ ln |x + 2| − ln |x + 1| = −
+ ln
x+1
x+1
x+1
And, take the limit to get −1 + ln(2)
4 5x2 + 3x − 2
dx
x3 + 2x2
Use partial fractions to get:
Z
Z
6.
−1 2
3
+ +
dx
2
x
x x+2
And integrate to get:
1
+ 2 ln(x) + 3 ln(x + 2)
x
5