Esercitazione guidata Analisi Matematica 1 - terza parte

Limiti, successioni e serie

1) Limite

1) limn->∞ log( cos  1⁄2n) + sin  1⁄3n⁄                    4√1 + 1⁄4n - 1log( cos 1⁄2n) = log (1 + (-1 + cos 1⁄2n)) ∼ -1 + cos 1⁄2n= -(1 - cos 1⁄2n) ∼ -1⁄2(1⁄2n)2 = -1⁄2 · 1⁄4nsen 1⁄3n ∼ 1⁄3nlog (cos 1⁄2n) + sen 1⁄3n = (sen 1⁄3n)      [                    log ( cos 1⁄2n) ⁄ sen 1⁄3n + 1     ]= (sen 1⁄3n)         [              log ( 1 + (-1 + cos 1⁄2n))⁄-1 + cos 1⁄2n                     +                    -1 + cos 1⁄2n⁄( 1⁄2n )2                   +1                                          +                                   sen 1⁄3n⁄1⁄3n          1⁄3n        ]= (sen 1⁄3n) ·log (1 + (-1 + cos 1⁄2n))⁄-1 + cos 1⁄2n · -1 + cos 1⁄2n⁄( 1⁄2n)2       · sen 1⁄3n⁄1⁄3n           ·  (             3⁄4         )        n    -1    = ( sen 1⁄3n)  ·1⁄3n     →

11) _limn→∞ ^{log(cos1/2ⁿ) + sin1/3ⁿ}/_{√⁴(1 + ¹/4ⁿ) - 1}log(cos^1/2ⁿ) = log(1 + (-1 + cos^1/2ⁿ))∼ -1 + cos^1/2ⁿ= - (1 - cos^1/2ⁿ) ∼ -¹/₂ ( ¹/_2ⁿ)² = -¹/₂ ⋅ ¹/_4ⁿsen^1/3ⁿ ∼ ¹/_3ⁿlog(cos^1/2ⁿ) + sen^1/3ⁿ = (sen^1/3ⁿ)[ ^{log(cos1/2ⁿ)}/_sen^1/3ⁿ + 1 ]= (sen^1/3ⁿ)[ ^{log(1 + (-1 + cos1/2ⁿ))}/_{-1 + cos^1/2ⁿ} ⋅ ^sen1/3ⁿ/_3ⁿ ⋅ ^(1/_2ⁿ)² ]log(1 + 2ⁿ)²ⁿ/_{1 - cos^2ⁿ 2ⁿ} → ¹/₂ se 2ⁿ → 0= ( ^sen1/3ⁿ )[ ^{log(1 + (-1 + cos1/2ⁿ))}/_{-1 + cos^1/2ⁿ} ⋅ ^sen1/3ⁿ/_1/3ⁿ→ 1D = (⁴√(1 + ¹/_4ⁿ) - 1) / ¹/_4ⁿ → ¹/₄(1 + 2n)^α - 1 / 2n → αN / D = (^{sen 1}/_3ⁿ)(^{sen 1}/_3ⁿ) / ¹/_3ⁿlog(1 + (-1 + cos ¹/_2¹)) . - 1 + cos ¹/_2¹- 1 + cos ¹/_2¹ (¹/_2¹)² . (³/₄)⁴ + 1 = ⁴√(1 + ¹/_4ⁿ - 1) / ¹/_4ⁿ . ¹/_4ⁿ(^{sen 1}/_3ⁿ) → 1⁴√(1 + ¹/_4ⁿ - 1 )( ¹/_4ⁿ → ¹/₄(⁴/₃)ⁿ → +∞

2) Limite

2) lim n → ∞[√(n³ + n) - √(n³ - 1)] / [√(n² + n) + √(n² - n)]

N = [√(n³ + n) - √(n³ - 1)] = (√(n³ + n) - √(n³ - 1)) . (√(n³ + n) + √(n³ - 1))= \(\frac{(n^3+n)- (n^3-1)}{\sqrt{n^3+n}+\sqrt{n^3-1}} = \frac{n+1}{\sqrt{n^5}(\frac{1+\frac{1}{n^2}}{n^2})+\sqrt{n^3}(1-\frac{1}{n^3})}\)= \(\frac{n+1}{n^{3/2}\sqrt{1+\frac{1}{n^2}}+\sqrt{1-\frac{1}{n^3}}}\sim \frac{n}{n^{3/2}\cdot 2} = \frac{1}{2\sqrt{n}} = \frac{1}{2\sqrt{n}}\)

D = \(\sqrt{n^2+\frac{\tg\frac{1}{n^2}}{n}} - n = \sqrt{n^2(1+\frac{1}{n^2}\tg\frac{1}{n^2})} - n\)= \(n\left(\sqrt{1+\frac{1}{n^2}\tg\frac{1}{n^2}} - 1\right)\sim x\cdot\frac{1}{2}\cdot\frac{1}{n^2}\tg\frac{1}{n^2}\)\(\sim\frac{1}{2}\cdot\frac{1}{n}\tg\frac{1}{n^2}\sim\frac{1}{2}\cdot\frac{1}{n}\cdot\frac{1}{n^2}=\frac{1}{2n^3}\)

\(\frac{N}{D} \sim\frac{\frac{1}{2\sqrt{n}}}{\frac{1}{2n^3}}=\frac{1}{\sqrt{n}}\cdot \frac{n^3}{\sqrt{n}} = \frac{n^3}{\sqrt{n}} \rightarrow +\infty\)

3) Limite

3) \(\lim_{n\to\infty}\frac{(2n)!}{2^{n^2}}\)

\(a_n = \frac{(2n)!}{2^{n^2}}\)

\(\frac{a_{n+1}}{a_n}=\frac{(2(n+1))!}{2^{(n+1)^2}}\cdot \frac{2^{n^2}}{(2n)!}=\ldots\)

(2(n+1))! = (2n)! (2n+1)(2n+2)2⁽ⁿ⁺¹⁾² = 2^n2+2n+1 = 2ⁿ² · 2²ⁿ⁺¹... = ^{(2n)! (2n+1)(2n+2)} · ^2x22ⁿ 2²ⁿ 2² (2n)!^x2= ^(2n+1)(2n+2) ~ ^{2n · 2n} = ⁴ⁿ² → 0 < 12²ⁿ · 2= D dn → 0.

Per casa

1) Limite

1) ^{lim nn}n→∞ _(n!)²

4) Limite

4) ^{lim nαn} α∈ℝn→∞ _(2n)!dn = ^nαn ≠ 0 ∀n∈ℕ, ∀α∈ℝ(2n)! (2n+1)(2n+2)?ⁿα = n+α→ ^(n+1)! · ^(2n)!.dn ^nαn(2(n+1))!= (n+1)^αn · (n+1)^α_{2^{x(2n)! (2n+1)(2n+2)^2}}· ^(2n)!. = (n+1)^αn · (n+1)^α· ^(n+1)αnn^α(1 + ¹)^α· ^(2+1)(2+2)2n dn^nαn n n= [ (1 + ¹/_n)ⁿ ]^α . n^α-2(n + ¹/_n)^α / (2 + ¹/_n) . (2 + ²/_n) → ¹/₄· α - 2 = 0 (α = 2)· α - 2 > 0 (α > 2)· α - 2 l = ^e2/₄ > 1 ⇨ a_n → ∞l = + ∞ ⇨ a_n → ∞l = 0 n → 0

Serie note e confronti

∑^∞_n=1 ¹/_n^p { converge se p > 1    diverge se p ≤ 1

∑^∞_n=0 xⁿ   { converge se |x|     diverge se x ≥ 1    indeterminata se x ≤ -1

a_n ≤ b_n ≤ c_n  ↓    ↓     ↓   (a)     (a)      (a)

0 ≤ a_n ≤ b_n   ↓  (b)

0 ≤ a_n ≤ b_n  ↓  ↓ +∞ +∞

5) Serie

5)\(\sum_{n=1}^{+\infty}\left(\frac{e^{\sin n}}{n^3+2}\right)\)

\(|\sin n| \leq 1\)

\(e^x\) crescente

\(\left(x_1 \lt x_2\right)\)

\(\left( e^{x_1} \lt e^{x_2} \right)\)

\(\Leftrightarrow -1 \leq \sin n \leq 1\)

\(0 \lt e^{\sin n} \lt e^1 = e\)

\(n^3 + 2 \gt n^3\)

\(\Rightarrow \frac{1}{n^3+2} \lt \frac{1}{n^3}\)

\(0 \leq \frac{e^{\sin n}}{n^3+2} \leq \frac{e}{n^3+2} \leq \frac{e}{n^3}\)

\(\sum_{n=1}^{\infty} \frac{e^{\sin n}}{n^3+2} \lt \sum_{n=1}^{\infty} \frac{e}{n^3} = e \sum_{n=1}^{\infty} \frac{1}{n^3}\) Converge

6) Serie

6)\(\sum_{n=0}^{+\infty}\log\left(\frac{n+3}{n+1}\right)\)

\(\log\left(\frac{n+3}{n+1}\right)=\log\left(\frac{(n+1)+2}{n+1}\right)=\log\left(1+\frac{2}{n+1}\right)\)

\(\sim \frac{2}{n+1} \sim \frac{2}{n}\)

\(\sum_{n=0}^{\infty}\log\left(\frac{n+3}{n+1}\right) \sim \sum_{n=1}^{\infty}\frac{2}{n}=2\sum_{n=1}^{\infty}\frac{1}{n}\)