Solutions to the third exam

1. This is a geometric series with the first two terms missing. Thus
∞
∑
n=2 		2
3n		 = ∞
∑
n=0 		2(1
3 		)n-2(1
3 		)0-2(1
3 		)1.
Applying the formula for a geometric series gives =
2
1-(1/3) 		-2- 2
3 		=3-2- 2
3 		= 1
3 		.

2. The basic error estimate for an alternating series is |S-SN| < bN+1.
So we want N such that bN+1<10-4. Thus we want
1
[(N+1)!]2		 < 1
104		.
Cross-multiplying shows that we want 104< [(N+1)!]2 , that is, 102< (N+1)!.
By trial-and-error, the least N that works is N=4.

3.
(a)
∑
5n
n!
We use the Ratio Test. We compute

lim
n→∞		 5n+1
(n+1)! 		·n!
5n		=
lim
n→∞		 5
n+1 		=0.
As 0 is less than 1, the series converges.

(b)
∑
2n-3
n2+7
We use the Limit Comparison Test, comparing to bn=n/n2=1/n.
We compute

lim
n→∞		 2n-3
n2+7 		÷1
n 		=
lim
n→∞		 2n2-3n
n2+7 		=
lim
n→∞		 2-(1/n)
1+(7/n2) 		=2.
Since 2 is not 0 or infinity, the test applies. Since Σbn=Σ(1/n)
diverges (it is a p-series with p=1), the given series also diverges.

(c)
∑
(n+1
2n+3 		)n
We use the Root Test. The nth root of an
is simply (n+1)/(2n+3).
So we compute

lim
n→∞		 n+1
2n+3 		=
lim
n→∞		 1+(1/n)
2+(3/n) 		= 1
2 		.
Since 1/2 is less than 1, the series converges.

(d)
∑
1
n(ln n)3
We use the Integral Test. Note that the series is only defined for n≥2,
as ln(1)=0. So we assume the series begins at n=2. To do the integral, let u = ln(x) so that du =(1/x)dx. Then
   
∫
1
x(ln x)3		 =    
∫
du
u3		=    
∫
u-3du =u-2/(-2) = -1
2u2		= -1
2(ln x)2		.
Thus the improper integral is

lim
t→∞		     t
∫
2 		dx
x(ln x)3		 =
lim
t→∞		 -1
2(ln x)2		 t
|
2		=
lim
t→∞		 -1
2(ln t)2		- -1
2(ln 2)2		= 1
2(ln 2)2
Thus the improper integral converges and so the given series also converges.

(e)
∑
n+1
2n+3
We use the Divergence Test (also called the nth term test). The limit of the nth term is

lim
n→∞		 n+1
2n+3 		=
lim
n→∞		 1+(1/n)
2+(3/n) 		= 1
2 		,
which is not zero. So the series diverges.

(e)
∑
cos2(n)
n2
We use the Direct Comparison Test. Note that 0 ≤cos2(n)≤1.
Thus
cos2(n)
n2		≤1
n2		=bn.
Now Σ(1/n2)
converges (it is a p-series with p=2), so the given series also converges.

4.(a) We first test the series in absolute value, namely,
Σ |an| =
∑
1
n1/2		,
which diverges (it is a p-series with p=1/2). Next we test the original series, which is an alternating series. So we check
1
(n+1)1/2		≤1
n1/2		  and  
lim
n→∞		 1
n1/2		=0
both of which are clearly true. So the alternating series converges. Thus this series converges conditionally.

4.(b) First note that this is not an alternating series. The terms go +, - , -, +, +,-, - , not +, -, +, -. We first test the series in absolute value, namely Σ|an|=Σcn.
We use the Ratio Test
cn+1
cn		= 1·3·5...·(2n-1)(2(n+1)-1)
2·7·12...·(5n-3)(5(n+1)-3) 		·2·7·12...·(5n-3)
1·3·5...·(2n-1) 		= 2n+1
5n+2
Taking limits gives

lim
n→∞		 2n+1
5n+2 		=
lim
n→∞		 2+(1/n)
5+(2/n) 		= 2
5 		.
As 2/5 is less than 1, the series converges absolutely.