2)
å1/(n(n+5)) = 1/5å(1/n-1/(n+5)) = 1/5(1-1/6+1/2-1/7+1/3-1/8+1/4-1/9+1/5-1/10+¼) = 1/5(1+1/2+1/3+1/4+1/5) = 137/300
3)
ò4¥ sec2(p/x)/x2dx = -tan(p/x)/p|4¥ = 1/p, therefore åsec2(p/n)/n2 is convergent.
4)
limn®¥ |an+1/an| = limn®¥(n+1)!/2n+1·2n/n! = limn®¥ (n+1)/2® ¥,
therefore ån!/2n is divergent.
5)
limn®¥| 2n+1(n+1)2/nn|1/n = limn®¥ 21+1/n(n+1)2/n/n = 0, therefore
å2n+1(n+1)2/nn is convergent.
6) limn®¥ |an|/(1/n) = 1, and å1/n
divergents implies that åan is not absolutely convergent by the limit
comparison test. However,
limn®¥ an = 0, (n+1)(n+2) > n(n+1)
implies an+1 < an
so åan is conditionally convergent by the alternating series test.
7) limn®¥ |an|/(1/n2) = 1 and å1/n2
is convergent so åan is absolutely convergent by the limit comparison
test.
8)
limn®¥ | an+1/an| = limn®¥ |x-5|n+1/2n+1·2n/|x-5|n = |x-5|/2 which must be less than 1 for convergence.
|x-5|/2< 1 will be true when 3 < x < 7. At
x = 3, åan = å1 which diverges. At
x = 7, åan = å(-1)n which diverges. Therefore, the interval of
convergence is 3 < x < 7.
9) f(4) = -8/3, f'(4) = 2/9, f(2)(4) = -4/27, f(3)(4) = 4/27
and f(4)(4) = -16/81 so we have
P4(x) = -8/3 + 2(x-4)/9 -4(x-4)2/(27·2!)+ 4(x-4)3/(27·3!)-16(x-4)4/(81·4!), so
P4(x) = -8/3 + 2(x-4)/9 -2(x-4)2/27+ 2(x-4)3/81- 2(x-4)4/243
10) If y = 1 +a1 x + a2x2 + a3x3 + a4x4 + ¼, then
xy = x + a1x2 + a2 x3 + a3 x4 + a4x5 + ¼ and
y' = a1 + 2a2x + 3a3 x2 + 4a4 x3 + 5a5 x4 + ¼. Therefore,
0 = y'-xy = a1 + (2a2-1)x + (3a3-a1)x2 + (4a4-a2)x3 + (5a5-a3)x4 +¼. So equating coefficients of like powers of x gives the following.
a1 = 0, 2a2-1 = 0 so a2 = 1/2, 3a3-a1 = 0 so a3 = 0, 4a4-a2 = 0
so a4 = 1/8 and 5a5-a3 = 0 so a5 = 0. Therefore, the first 5 terms are
y = 1 + x2/2 + x4/8.