Notes about slide 7

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University of Toronto – MAT137Y1 – LEC0501

Calculus!

Notes about slide 7

Jean-Baptiste Campesato March 4

^th

, 2019

For each𝑛 ∈ ℕ_>0, we define

𝑟_𝑛= the smallest power of2that is greater than or equal to𝑛 Consider the series𝑆 =

∞

∑𝑛=1

1 𝑟_𝑛. 1. Compute𝑟₁through𝑟₈.

𝑟₁= 1,𝑟₂= 2,𝑟₃= 4,𝑟₄= 4,𝑟₅= 8,𝑟₆= 8,𝑟₇= 8,𝑟₈= 8.

2. Compute the partial sums𝑆₁, 𝑆₂, 𝑆₄, 𝑆₈of the series𝑆.

𝑆₁=

1

∑𝑛=1

1 𝑟_𝑛 = 1

𝑟₁ = 1

𝑆₂=

2

∑𝑛=1

1 𝑟_𝑛 = 1

𝑟₁ + 1

𝑟₂ = 1 + 1 2 = 3

2

𝑆₄=

4

∑𝑛=1

1 𝑟_𝑛 =

2

∑𝑛=1

1 𝑟_𝑛+

4

∑𝑛=3

1

𝑟_𝑛 = 𝑆₂+ 1 𝑟₃+ 1

𝑟₄ = 𝑆₂+1 4 +1

4 = 𝑆₂+1 2 = 3

2 +1 2 = 2

𝑆₈=

8

∑𝑛=1

1 𝑟_𝑛 =

4

∑𝑛=1

1 𝑟_𝑛+

8

∑𝑛=5

1

𝑟_𝑛 = 𝑆₄+1 8+ 1

8+ 1 8+ 1

8 = 𝑆₄+1

2 = 2 +1 2 = 5

2

It seems that𝑆₂𝑘+1= 𝑆₂𝑘+ 1

2, allowing us to conjecture that

∀𝑘 ∈ ℕ, 𝑆₂𝑘 = 1 +𝑘 2

The above claim is true and will be useful for the next question, so let’s prove it by induction.

Base case, for𝑘 = 0:𝑆₂0= 𝑆₁= 1 = 1 +⁰

2.

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2 Notes about slide 7

Induction step:

We assume that𝑆₂𝑘 = 1 +^𝑘₂ for some𝑘 ∈ ℕand we want to show that𝑆₂𝑘+1= 1 + ^𝑘+1₂ . First, notice that

𝑆₂𝑘+1=

2^𝑘+1

∑𝑛=1

1 𝑟_𝑛 =

2^𝑘

∑𝑛=1

1 𝑟_𝑛+

2^𝑘+1 𝑛=2∑^𝑘+1

1

𝑟_𝑛 = 𝑆₂𝑘+

2^𝑘+1 𝑛=2∑^𝑘+1

1

𝑟_𝑛 = 1 + 𝑘 2 +

2^𝑘+1 𝑛=2∑^𝑘+1

1 𝑟_𝑛

where the last equality comes from the induction hypothesis.

Then, notice that the last sum has(2^𝑘+1) − (2^𝑘+ 1) + 1 = 2^𝑘+1− 2^𝑘= 2^𝑘(2 − 1) = 2^𝑘terms and moreover, for all𝑛 = 2^𝑘+ 1, 2^𝑘+ 2, … , 2^𝑘+1, we have2^𝑘< 𝑛 ≤ 2^𝑘+1, hence𝑟_𝑛= 2^𝑘+1.

Therefore

𝑆₂𝑘+1= 1 + 𝑘

2 + 2^𝑘 1

2^𝑘+1 = 1 + 𝑘 2 +1

2 = 1 +𝑘 + 1 2 3. Compute𝑆 =

∞

∑𝑛=1

1 𝑟_𝑛.

We denote the𝑘-th partial sum of𝑆by𝑆_𝑘=

𝑘

∑𝑛=1

1 𝑟_𝑛. Let𝑘 ∈ ℕ_>0, then

𝑆_𝑘+1=

𝑘+1

∑𝑛=1

1 𝑟_𝑛 =

⎛⎜

⎜⎝

𝑘

∑𝑛=1

1 𝑟_𝑛

⎞⎟

⎟⎠ + 1

𝑟_𝑘+1 = 𝑆_𝑘+ 1 𝑟_𝑘+1 > 𝑆_𝑘 Hence the sequence(𝑆_𝑘)_𝑘≥1is increasing.

Therefore, either it is bounded from above and then convergent or it is not bounded from above and then it is divergent to+∞.

But we know that lim

𝑘→+∞𝑆₂𝑘 = lim

𝑘→+∞(1 +𝑘

2 )= +∞.

So(𝑆_𝑘)_𝑘is not bounded from above: for any𝑀 ∈ ℝ, there exists𝑙 ∈ ℕsuch that𝑆_𝑘 > 𝑀 with𝑘 = 2^𝑙.

Hence, since(𝑆_𝑘)_𝑘is increasing and not bounded from above, we get that lim

𝑘→+∞𝑆_𝑘= +∞.

We proved that

∞

∑𝑛=1

1

𝑟_𝑛 = +∞.

4. Compute𝐻 =

∞

∑𝑛=1

1 𝑛. Hint:“Compare”𝐻and𝑆.

By definition of𝑟_𝑛, we know that, for any𝑛 ∈ ℕ_>0, we have𝑟_𝑛≥ 𝑛 > 0and therefore_𝑟¹

𝑛 ≤ ¹_𝑛. Hence, for any𝑘 ∈ ℕ_>0, we have

𝑘

∑𝑛=1

1 𝑟_𝑛 ≤

𝑘

∑𝑛=1

1 𝑛. Since lim

𝑘→+∞

𝑘

∑𝑛=1

1

𝑟_𝑛 = +∞, we get by comparison that lim

𝑘→+∞

𝑘

∑𝑛=1

1 𝑛 = +∞.

We proved that

∞

∑𝑛=1

1 𝑛 = +∞.