MA 1505 Mathematics I

Tutorial 3 Solutions

1. (a) Let un = (−1)n (x+2)n

n .

limn→∞

∣∣∣∣un+1

un

∣∣∣∣ = limn→∞

∣∣∣∣(x+ 2)n+1

n+ 1· n

(x+ 2)n

∣∣∣∣ = |x+ 2|.

By ratio test, the power series is convergence in |x+ 2| < 1.

So the radius of convergence is 1.

(b) Let un = (3x−2)nn .

limn→∞

∣∣∣∣un+1

un

∣∣∣∣ = limn→∞

∣∣∣∣(3x− 2)n+1

n+ 1· n

(3x− 2)n

∣∣∣∣ = |3x− 2|.

By ratio test, the power series is convergence in |3x− 2| < 1⇒ |x− 23 | <

13 .

So the radius of convergence is1

3.

(c) Let un = (−1)n(4x+ 1)n.

limn→∞

∣∣∣∣un+1

un

∣∣∣∣ = limn→∞

∣∣∣∣(4x+ 1)n+1

(4x+ 1)n

∣∣∣∣ = |4x+ 1|.

By ratio test, the power series is convergence in |4x+ 1| < 1⇒ |x+ 14 | <

14 .

So the radius of convergence is1

4.

(d) Let un = (3x)n

n! .

limn→∞

∣∣∣∣un+1

un

∣∣∣∣ = limn→∞

∣∣∣∣ (3x)n+1

(n+ 1)!· n!

(3x)n

∣∣∣∣ = limn→∞

∣∣∣∣ 3x

n+ 1

∣∣∣∣ = 0.

Since the limit is less than 1 for any x, so the radius of convergence is ∞.

(e) Let un = (nx)n.

limn→∞

∣∣∣∣un+1

un

∣∣∣∣ = limn→∞

∣∣∣∣((n+ 1)x)n+1

(nx)n

∣∣∣∣ = limn→∞

∣∣∣∣(1 +1

n

)n

(n+ 1)x

∣∣∣∣ =∞

for all x 6= 0.

So the radius of convergence is 0.

(f) Let un = (4x−5)2n+1

n3/2 .

limn→∞

∣∣∣∣un+1

un

∣∣∣∣ = limn→∞

∣∣∣∣∣(4x− 5)2n+3

(n+ 1)3/2· n3/2

(4x− 5)2n+1

∣∣∣∣∣ = |4x− 5|2.

By ratio test, the power series is convergence in |4x− 5|2 < 1⇒ |4x− 5| < 1⇒ |x− 54 | <

14 .

So the radius of convergence is1

4.

MA1505 Tutorial 3 Solutions

2. The first term of the geometric series is a = 1 and the common ratio is r = −(x− 3)

2.

So the sum of the series is

a

1− r=

1

1 + (x− 3)/2)=

2

x− 1

provided

∣∣∣∣(x− 3)

2

∣∣∣∣ < 1⇒ 1 < x < 5.

3. (a)

x

1− x= x

(1

1− x

)= x

∞∑n=0

xn =∞∑n=0

xn+1

(b) Let f(x) = 1x2 .

Then f ′(x) = − 2

x3, f ′′(x) =

3!

x4, . . . and in general f (n)(x) = (−1)n

(n+ 1)!

xn+2.

So f (n)(1) = (−1)n(n+ 1)!.

The Taylor series of f at x = 1 is thus:

f(x) =∞∑n=0

fn(1)

n!(x− 1)n =

∞∑n=0

(−1)n(n+ 1)(x− 1)n.

(c)

x

1 + x=

(1 + x)− 1

1 + x= 1− 1

1 + x= 1− 1

1 + (x+ 2)− 2

= 1 +1

1− (x+ 2)= 1 +

∞∑n=0

(x+ 2)n = 2 +∞∑n=1

(x+ 2)n

4. Given f(x) = sinx, compute the following:

f(x) = sinx f(0) = sin 0 = 0

f (1)(x) = cosx f (1)(0) = cos 0 = 1

f (2)(x) = − sinx f (2)(0) = − sin 0 = 0

f (3)(x) = − cosx f (3)(0) = − cos 0 = −1

f (4)(x) = sinx

Taylor polynomial at x = 0 of order 3 is

P3(x) =3∑

k=0

f (k)(0)

k!xk = 0 +

1

1!x +

0

2!x2 +

−1

3!x3 = x− 1

6x3.

2

MA1505 Tutorial 3 Solutions

P3(0.1) approximates sin 0.1 as follows:

P3(0.1) = 0.1− 1

6(0.1)3 ≈ 0.09983.

In using P3(x) to approximate sinx the error incurred is

R3(x) =f (4)(c)

4!x4 for some c between 0 and x.

Thus, magnitude of error incurred is

|R3(0.1)| =

∣∣∣∣sin c4!(0.1)4

∣∣∣∣ ≤ (0.1)4

4!since |sin θ| ≤ 1 for any θ

< 10−5.

5. (i) We have

xex = x

∞∑n=0

xn

n!=

∞∑n=0

xn+1

n!

Note that the radius of convergence of this power series is infinite. Therefore,we can integrate

both sides from 0 to 1.∫ 1

0xex dx =

∞∑n=0

∫ 1

0

xn+1

n!dx =

∞∑n=0

1

n!(n+ 2).

On the other hand, ∫ 1

0xex dx = [xex]10 −

∫ 1

0ex dx = e− (e− 1) = 1.

So we conclude∞∑n=0

1

n!(n+ 2)= 1.

(ii) We have

ex − 1

x=

∑∞n=1

xn

n!

x=

∞∑n=1

xn−1

n!=

∞∑n=0

xn

(n+ 1)!.

Note that the radius of convergence of this power series is infinite.

Differentiate both sides with respect to x, we have

xex − (ex − 1)

x2=∞∑n=1

nxn−1

(n+ 1)!=∞∑n=0

(n+ 1)xn

(n+ 2)!.

The result now follows by setting x = 1.

3

MA1505 Tutorial 3 Solutions

6. First we calculate

12

∫ 10 t

n−1 (1− t)2 dt

= 12

∫ 10 t

n−1 (1− 2t+ t2)dt

= 12

[1n t

n − 2n+1 t

n+1 + 1n+2 t

n+2]10

= 1n(n+1)(n+2)

Next we apply this result to n = 1, 3, 5, 7, · · ·, we find that

11·2·3 + 1

3·4·5 + 15·6·7 + 1

7·8·9 + · · ·

= 12

∫ 10 t

1−1 (1− t)2 dt+ 12

∫ 10 t

3−1 (1− t)2 dt+ 12

∫ 10 t

5−1 (1− t)2 dt+ 12

∫ 10 t

7−1 (1− t)2 dt+ · · ·

= 12

∫ 10 (1− t)2

(1 + t2 + t4 + t6 + · · ·

)dt

= 12

∫ 10 (1− t)2

(1

1−t2

)dt

= 12

∫ 10

1−t1+tdt

= −12

∫ 10

1+t−21+t dt

= −12

∫ 10

(1− 2

1+t

)dt

= ln 2− 12

4