Esercitazione guidata Analisi Matematica 1 - quarta parte

1) Serie con termine ²³ⁿ/_(3n)!

∑_n=1^{∞ 23n}/_(3n)!

lim_n→∞ ²³ⁿ/_(3n)!

a_n = ²³ⁿ/_(3n)! ≠ 0 ∀ n ∈ ℕ

a_n+1/_an = ^23(n+1)/_(3(n+1))! · (3n)! / 2³ⁿ = ^23n·23/[(3n)!·(3n+1)(3n+2)(3n+3)] = ⁸/_27n³ → 0 < 1 ⇒ a_n → 0 ⇒ la serie converge

2) Serie con termine ^{n3n·sin(1/n2)}/_(3n)!

∑_n=1^{∞ n3n·sin(1/n2)}/_(3n)! = a_n

a_n ≠ 0 ∀ n ∈ ℕ

a_n+1/a_n = ^{(n+1)3(n+1)·sin(1/(n+1)2)}/(3(n+1))! · (3n)!/(n³ⁿ·n·sin(1/n²)) = [(n+1)/n]³ⁿ · [sin(1/(n+1)²)/sin(1/n²)] · [1/(3n+1)·(3n+2)·(3n+3)] ≈ ...

1) Serie con termine ²³ⁿ⁄_(3n)!

∑_n=1^{∞     23n}⁄_(3n)!

lim_n→∞   ²³ⁿ⁄_(3n)!

a_n   ²³ⁿ⁄_(3n)!   ≠ 0   ∀n∈&naturals;

^a_n+1⁄_an = ^23(n+1)⁄_(3(n+1))! · ^(3n)!⁄23n  = ^{23 · 2}⁄_{(3n)! · (3n+1)(3n+2)(3n+3)} ·  ≈ ⁸⁄_27n³ → 0 < 1   →   a_n → 0  ⇒ la serie converge

2) Serie con termine ^{n3nsin(⁄n2)}⁄_(3n)!

∑_n=1^{∞   n3nsin(⁄n2)}⁄_(3n)!   = a_n

a_n ≠ 0   ∀n∈&naturals;

^a_n+1⁄_an = ^{(n+1)3(n+1) sin(⁄(n+1)2)}⁄_(3(n+1)!)! · ^{(3n)! n sin(⁄n2)}⁄  = ^{(n+1)3(n+1)3 sin(⁄(n+1)2)}⁄_{(3n)! · (3n+1)(3n+2)(3n+3)}    sin(⁄n²) ·  = ^(n+1)⁄n3n · ^{sin(⁄(n+1)2)}⁄_sin(⁄n²)·    ^{(3n+1)(3n+2)(3n+3)}⁄_(n+1)³  ≈ ...

sin_(1/(n+1)²) ∼ ₁/⁽ⁿ⁺¹⁾² {sin_(1/n²) ∼ ₁/ⁿ²}

sin_(1/(n+1)²) sin_(1/n²) ∼ ⁿ²/_(n+1)² ∼ _(n+1)/ⁿ³ⁿ ⋅ ⁿ²/_(n+1)² ⋅ _(n+1)³/²⁷ⁿ³ ∼ _(n+1)/ⁿ³ⁿ ⋅ _n/²⁷ⁿ = ₁/²⁷[_{(1 + 1)}/ⁿ]³ → e³/₂₇ < 1

(_n+1/ⁿ)³ⁿ = [(_{1 + 1}/ⁿ)ⁿ]³ → e³

=> a_n → 0 => la serie converge.

(_{1 + 1}/ⁿ)ⁿ2_1ⁿ = 1

lim_n→∞ 1ⁿ

f ∼ g

h ∼ t

f^h ∼ g^t

3) Serie con termine ⁿ²²ⁿ/_nⁿ²

∑_n=1^{∞ n22n}/_nⁿ²

a_n = ⁿ²²ⁿ/_nⁿ² = n^{2 (3/n)}√ⁿ2_n = √ⁿ(ⁿ²²ⁿ/_n²) = ²ⁿ/_{n sqrtⁿ²} → 0 < 1

\( \frac{n^2 \cdot 2^n}{n^n} \) = \( \frac{\sqrt[n]{n^2 \cdot n^2}}{\sqrt[n]{n^n}} \) = \( \frac{\sqrt[n]{n^2} \cdot 2}{n} \) \( \Rightarrow a_n \to 0

4) Serie con termine \(\frac{1}{3^{2n}}\left(\frac{n}{n-3}\right)^{n^2}\)

\( \sum_{n=4}^{\infty} \frac{1}{3^{2n}} \left( \frac{n}{n-3} \right)^{n^2} \)

\( a_n = \frac{1}{3^{2n}} \left( \frac{n}{n-3} \right)^{n^2} > 0 \)

\( \forall n \geq 4 \)

\( \sqrt[2] a_n = \sqrt[n]{\frac{1}{3^{2n}} \left( \frac{n}{n-3} \right)^{n^2}} = \frac{1}{3^2} \left( \frac{n}{n-3} \right)^n \)

\( \frac{1}{9} \left( \frac{n}{n-3} \right)^n \)

\( \frac{1}{9} \left( \frac{n}{n-3} \right)^n \)

\( \frac{1}{9} (e^{-3})^{-1} = \frac{e^3}{9} > 1 \)

\( \Rightarrow a_n \to \infty \Rightarrow \) la serie diverge.

5) Serie con termine \(n^2\log\left(\cos\left(\frac{1}{2^n}\right)\right)\)

\(\sum_{{n=0}}^{\infty} n^2 \log \left( \cos \left( \frac{1}{2^n} \right) \right)\)

\(n^2 \log \left( \cos \frac{1}{2^n} \right) = n^2 \log \left( 1 + \left( -1 + \cos \left( \frac{1}{2^n} \right) \right) \right)\)

\(\log (1 + a_n ) \sim a_n \quad a_n \to 0\)

\(\sim n^2 \left( -1 + \cos \left( \frac{1}{2^n} \right) \right)\)

\(1 - \cos a_n \sim \frac{a_n^2}{2} \qquad a_n \to 0\)

\(\sim n^2 \left( -\frac{1}{2} \left( \frac{1}{2^n} \right)^2 \right) = -\frac{1}{2} \frac{n^2}{2^{2n}} \to 0\)

\(\sqrt{\frac{n^2}{2^{2n}}} = \frac{n^n}{2^n}\) \(\to \frac{1}{4} < 1\)

\(\Rightarrow\) la serie converge

6) Criterio di Leibniz

\(\sum_{{n=1}}^{\infty} (-1)^n \left( \frac{2n - 1}{n^2} \right)\)

\(2n > 0 \quad \forall n \in \mathbb{N}\)

\(2n \to 0\)

\(2n\) decrescente

\(a_{n+1} \leq a_n\)

\(\frac{2(n+1) - 1}{(n+1)^2} \leq \frac{2n - 1}{n^2}\)

\(\Leftrightarrow \frac{2n + 1}{(n+1)^2} \leq \frac{2n - 1}{n^2}\)

Criterio di Leibniz

Se \((a_n)\):

• \(a_n > 0\)
• \(a_n \to 0\)
• \(a_n\) decrescente

\(\Rightarrow \sum_{{n=0}}^{\infty} (-1)^n a_n\) è convergente

(<=>) n²(2n+1) ≤ (n+1)²(2n-1)

(<=>) 2n³ + n² ≤ (n² + 2n + 1)(2n - 1)

(<=>) 2n³ + n² ≤ 2n³ - n² + 4n² - 2n + 2n² - 1

(<=>) 2n³ + n² ≤ 2n³ + 3n² - 1

(<=>) 3n² - 1 - n²? 0

(<=>) 2n² - 1? 0   ∀n ∈ IN

Per il Criterio di Leibniz la serie converge.

7) Serie con parametro α

∑ _n=0 ^∞ (³√n² - n + 1 - ³√n² - αn)   α ∈ ℝ

a_n = ³√n² - n + 1 - ³√n² - αn = ³√n² - n + 1(³√n² - n + 1 - ³√n² - αn)

n² - n + 1

n² - αn

= ³√n² - n + 1 - 1 = ³√n² + n - αn + αn - 1

= ³√(n² - αn) + (αn - n + 1) - 1

n² - αn = ³√1 + ^{αn - n + 1} - 1

n² - αn ∼ ¹₃ • (^α_-1)n + 1 ∼αx

a_n = ³√_{n² - αn} [^{3 n2-n+1}/_n²-αn -1] ≈ ³√_{n² - αn} · ¹/₃ · ^{(α-1)n + 1}/_{n² - αn}

= ¹/₃ · ^{(α-1)n + 1}/_{(n² - αn)^{1 - 1/3}} = ¹/₃ · ^{(α-1)n + 1}/_{(n² - αn)^2/3}

= ¹/₃ · ^{(α-1)n + 1}/_{n²(1 - α/n)^2/3} = ¹/₃ · ^{(α-1)n + 1}/_{n^4/3(1 - α/n)^2/3} ≈ ...

· Se   α-1 = 0   ⇒   α = 1

¹/₃ · ¹/_{n^4/3(1 - 1/n)^2/3} ≈ ¹/₃ ¹/_n^4/3

lim_n→∞ (1 - 1/n)^2/3 = 1     ⇒   (λ - 1/n)^2/3 ≈ 1

∑_n=1^∞ a_n ≈ ∑_n=1^{∞ 1}/_n^4/3          conv.

∑ ¹/_n^p   p>1   conv      p ≤ 1   div.

· Se   α-1 ≠ 0   ⇒   α ≠ 1