Solutions for slide 6

University of Toronto – MAT137Y1 – LEC0501

Calculus!

Solutions for slide 6

Jean-Baptiste Campesato April 3

, 2019

Disclaimer: those arequick-and-dirtynotes written just after the class, so it is very likely that they contain some mistakes/typos…

5.

+∞

∑𝑛=1

𝑛 3^𝑛

Notice that

+∞

∑𝑛=1

𝑛 3^𝑛 =

+∞

∑𝑛=1𝑛 (1 3 )

𝑛

= 𝑆₅(1

3 ) where 𝑆₅(𝑥) =

+∞

∑𝑛=1

𝑛𝑥^𝑛 has a radius of conver- gence equal to1(check it).

Hence for𝑥 ∈ (−1, 1)we have:

𝑆₅(𝑥) =

+∞

∑𝑛=1

𝑛𝑥^𝑛= 𝑥

+∞

∑𝑛=1

𝑛𝑥^𝑛−1= 𝑥

+∞

∑𝑛=1

𝑑

𝑑𝑥 (𝑥^𝑛) = 𝑥 𝑑 𝑑𝑥 (

+∞

∑𝑛=1

𝑥^𝑛

)= 𝑥𝑑 𝑑𝑥 (

𝑥

1 − 𝑥 )= 𝑥 (1 − 𝑥)²

Hence

+∞

∑𝑛=1

𝑛

3^𝑛 = 𝑆₅(1 3 )= 3

4

6.

+∞

∑𝑛=1

𝑛² 3^𝑛

For this question, it is interesting to introduce𝑆₆(𝑥) =

+∞

∑𝑛=1

𝑛²𝑥^𝑛whose radius of convergence is1(check it).

Hence for𝑥 ∈ (−1, 1)we have:

𝑆₆(𝑥) =

+∞

∑𝑛=1

𝑛²𝑥^𝑛= 𝑥

+∞

∑𝑛=1

𝑛 ⋅ 𝑛𝑥^𝑛−1= 𝑥

+∞

∑𝑛=1

𝑛 𝑑

𝑑𝑥 (𝑥^𝑛) = 𝑥 𝑑 𝑑𝑥 (

+∞

∑𝑛=1

𝑛𝑥^𝑛

)= 𝑥𝑑 𝑑𝑥𝑆₅(𝑥)

= 𝑥𝑑 𝑑𝑥 (

𝑥

(1 − 𝑥)²)= 𝑥(1 − 𝑥)²+ 2𝑥(1 − 𝑥)

(1 − 𝑥)⁴ = 𝑥1 − 𝑥 + 2𝑥 (1 − 𝑥)³

= 𝑥(1 + 𝑥) (1 − 𝑥)³ Hence

+∞

∑𝑛=1

𝑛²

3^𝑛 = 𝑆₆(1 3 )= 3

2

2 Solutions for slide 6

7. 𝑆₇(𝑥) =

+∞

∑𝑛=0

𝑥^𝑛 (𝑛 + 2)𝑛!

Notice that the radius of convergence of𝑆₇is+∞(check it).

Hence, for𝑥 ∈ ℝ ⧵ {0}: (I am going to divide by𝑥so I’ll treat the case𝑥 = 0after).

𝑆₇(𝑥) =

+∞

∑𝑛=0

𝑥^𝑛

(𝑛 + 2)𝑛! = 1 𝑥²

+∞

∑𝑛=0

1 𝑛!

𝑥^𝑛+2 𝑛 + 2 = 1

𝑥²

+∞

∑𝑛=0

1 𝑛! ∫

𝑥 0

𝑡^𝑛+1𝑑𝑡 = 1 𝑥²∫

𝑥 0

𝑡

+∞

∑𝑛=0

1 𝑛!𝑡^𝑛𝑑𝑡

= 1 𝑥²∫

𝑥 0

𝑡𝑒^𝑡𝑑𝑡

= 1

𝑥²([𝑡𝑒^𝑡]^𝑥0−

∫

𝑥 0

𝑒^𝑡𝑑𝑡) Integration by parts

= 1

𝑥²(𝑥𝑒^𝑥− 𝑒^𝑥+ 1)

And𝑆₇(0) = ¹₂ (that’s simply the constant term of the series: all the𝑥^𝑛vanishes at𝑥 = 0 except𝑥⁰which is a notation for1).

Hence𝑆₇(𝑥) = {

𝑥𝑒^𝑥−𝑒^𝑥+1

𝑥² if𝑥 ≠ 0

1

2 otherwise

8.

+∞

∑𝑛=0

(−1)^𝑛 (2𝑛 + 1)3^𝑛

Again, notice that

+∞

∑𝑛=0

(−1)^𝑛 (2𝑛 + 1)3^𝑛 =

+∞

∑𝑛=0

(−1)^𝑛 2𝑛 + 1 (

1

√3 )

2𝑛

so we introduce𝑆₈(𝑥) =

+∞

∑𝑛=0

(−1)^𝑛 2𝑛 + 1𝑥^2𝑛 whose radius of convergence is 1 (check it).

Hence, for𝑥 ∈ (−1, 1) ⧵ {0}(I’m going to divide by𝑥so I avoid the case𝑥 = 0which is not useful to answer this question).

𝑆₈(𝑥) =

+∞

∑𝑛=0

(−1)^𝑛

2𝑛 + 1𝑥^2𝑛= 1 𝑥

+∞

∑𝑛=0

(−1)^𝑛

2𝑛 + 1𝑥^2𝑛+1= 1 𝑥

+∞

∑𝑛=0

(−1)^𝑛

∫

𝑥 0

𝑡^2𝑛𝑑𝑡

= 1 𝑥 ∫

𝑥 0

+∞

∑𝑛=0

(−1)^𝑛𝑡^2𝑛𝑑𝑡 = 1 𝑥 ∫

𝑥 0

+∞

∑𝑛=0

(−𝑡²)^𝑛𝑑𝑡 = 1 𝑥 ∫

𝑥 0

1

1 + 𝑡²𝑑𝑡 = arctan𝑥 𝑥

And

+∞

∑𝑛=0

(−1)^𝑛

(2𝑛 + 1)3^𝑛 = 𝑆₈ (

1

√3 )= √3arctan (

1

√3 )= √3𝜋 6 = 𝜋

2√3

MAT137Y1 – LEC0501 – J.-B. Campesato 3

9.

+∞

∑𝑛=0

(−1)^𝑛𝑛 + 1 (2𝑛)!2^𝑛 First notice that

+∞

∑𝑛=0

(−1)^𝑛𝑛 + 1 (2𝑛)!2^𝑛=

+∞

∑𝑛=0

(−1)^𝑛𝑛

(2𝑛)! (√2)^2𝑛+

+∞

∑𝑛=0

(−1)^𝑛

(2𝑛)! (√2)^2𝑛

The second sum is cos(√2).

The first sum is equal to𝑓 (√2)where𝑓 (𝑥) =

+∞

∑𝑛=0

(−1)^𝑛𝑛

(2𝑛)! 𝑥^2𝑛 has a radius of convergence equal to+∞(check it).

Then for𝑥 ∈ ℝ,(notice that for𝑛 = 0, the coefficient is zero) 𝑓 (𝑥) =

+∞

∑𝑛=0

(−1)^𝑛𝑛 (2𝑛)! 𝑥^2𝑛= 𝑥

2

+∞

∑𝑛=0

(−1)^𝑛

(2𝑛)!2𝑛𝑥^2𝑛−1 = 𝑥 2

+∞

∑𝑛=0

(−1)^𝑛 (2𝑛)!

𝑑 𝑑𝑥 (𝑥^2𝑛)

= 𝑥 2

𝑑 𝑑𝑥 (

+∞

∑𝑛=0

(−1)^𝑛 (2𝑛)!𝑥^2𝑛

)= 𝑥 2

𝑑

𝑑𝑥(cos(𝑥)) = −𝑥sin𝑥 2

Then

+∞

∑𝑛=0

(−1)^𝑛𝑛

(2𝑛)! (√2)^2𝑛= 𝑓 (√2) = −√2

2 sin(√2).

And +∞

∑𝑛=0

(−1)^𝑛𝑛 + 1

(2𝑛)!2^𝑛=cos(√2) −√2

2 sin(√2)

Notice that if you didn’t think about the√2that’s not a big deal here, indeed:

+∞

∑𝑛=0

(−1)^𝑛𝑛

(2𝑛)! 2^𝑛= 𝑔(2) where, if𝑥 > 0,

𝑔(𝑥) =

+∞

∑𝑛=0

(−1)^𝑛𝑛 (2𝑛)! 𝑥^𝑛

= 𝑥

+∞

∑𝑛=0

(−1)^𝑛 (2𝑛)!𝑛𝑥^𝑛−1

= 𝑥

+∞

∑𝑛=0

(−1)^𝑛 (2𝑛)!

𝑑 𝑑𝑥 (𝑥^𝑛)

= 𝑥𝑑 𝑑𝑥 (

+∞

∑𝑛=0

(−1)^𝑛 (2𝑛)!𝑥^𝑛

)

= 𝑥𝑑

𝑑𝑥 (cos(√𝑥))

= −

𝑥sin(√𝑥) 2√𝑥