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EXERCISES - SERIES


+∞Σk=0 k/(k2+1); necessary condition limk→∞ ak = limk→∞ k/(k2+1) = 0; it is a positive term series, ak = k/(k2+1) ≥ 0 ∀ k ∈ ℕ; limk→∞ ak/bk = l ≠ 0, then +∞Σk=0 ak < +∞ ⇔ +∞Σk=0 bk < +∞, the two series have the same behavior; k/(k2+1) ~ k/k2 = 1/k ⇔ limk→∞ (k/(k2+1))/(1/k) = 1; +∞Σk=1 1/k = +∞ ⇒ +∞Σk=1 k/(k2+1) = +∞

+∞Σk=1 1/(3k-k); necessary condition limk→∞ 1/(3k-k) = limk→∞ 1/(3k(1-k/3k)) = 0; limk→∞ k/3k = 0 ⇒ 3k grows faster than k ⇒ 3k-k > 0, it is a positive term series; (3k-k) ~ 3k ⇔ 1/(3k-k) ~ 1/3k; +∞Σk=1 1/3k = +∞Σk=1 (1/3)k < +∞ ⇒ +∞Σk=1 1/(3k-k) < +∞

+∞Σk=1 ln(k)/k3; limk→∞ ln(k)/k3 = 0; ln(k) ≥ 0 ∀ k ∈ ℕ, k ≥ 1, it is a positive term series; ln(k)/k3 ~ ?/k3; according to the second part of the asymptotic comparison criterion, if limk→∞ ak/bk = 0, then +∞Σk=0 bk < +∞ ⇒ +∞Σk=0 ak < +∞; limk→∞ (ln(k)/k3)/(1/k2) = limk→∞ (ln(k)/k3)k2 = 0; +∞Σk=1 1/k2 < +∞ ⇒ +∞Σk=1 ln(k)/k3 < +∞; +∞Σk=1 1/kα = {< +∞, α > 1; = +∞, α ≤ 1}

+∞Σk=1 ln(k)/k2 < +∞; limk→∞ ln(k)/k2 = 0; ln(k) ≥ 0 ∀ k ∈ ℕ, k ≥ 1, it is a positive term series; ln(k)/k2 ~ ?/k2; according to the second part of the asymptotic comparison criterion, if limk→∞ ak/bk = 0, then +∞Σk=0 bk < +∞ ⇒ +∞Σk=0 ak < +∞; limk→∞ (ln(k)/k2)/(1/k3/2) = limk→∞ ln(k)/k1/2 = 0; +∞Σk=1 1/k3/2 < +∞ ⇒ +∞Σk=1 ln(k)/k2 < +∞; limk→∞ ln(k)/kα = 0 ∀ α > 0; +∞Σk=1 1/kα = {< +∞, α > 1; = +∞, α ≤ 1}

+∞Σk=1 k!/kk; limk→∞ k!/kk = 0, the necessary condition is verified; it is a positive term series; k!/kk ~ ?/kk; it is possible to use the Stirling's approximation or Stirling's formula, but the most logical way is to use the ratio criterion; ak = k!/kk, limk→∞ ak+1/ak = limk→∞ ((k+1)!/(k+1)k+1)/(k!/kk) = limk→∞ (k!(k+1)/(k+1)k(k+1))(kk/k!) = limk→∞ kk/(k+1)k = limk→∞ (k/(k+1))k = limk→∞ 1/((k+1)/k)k = limk→∞ 1/(1+1/k)k = 1/e < 1, according to the ratio criterion, the series converges

+∞Σk=1 xk/k, x ∈ ℝ; considering the absolute convergence, if the series absolutely converges, then it converges; +∞Σk=1 |xk/k| = +∞Σk=1 |x|k/k, it is a positive term series, so it is possible to apply the ratio criterion; |xk| = |x⋅x⋅...⋅x| = |x|⋅|x|⋅...⋅|x| = |x|k; limk→∞ k|x|k/k = limk→∞ |x|/kk = |x| < 1 ⇒ +∞Σk=1 |xk/k| converges, then +∞Σk=1 xk/k absolutely converges and therefore converges; |x| = 1, case of doubt; |x| > 1, +∞Σk=1 |xk/k| diverges, that is, it does not absolutely converge, therefore the starting series cannot be defined; |x| > 1 ⇔ x < -1 and x > 1; x > 1, limk→∞ xk/k = +∞, the necessary condition is not verified, the series does not converge; x < -1, limk→∞ xk/k = limk→∞ ((-1)(-x))k/k, limk→∞ ((-1)k(-x)k)/k ∄, the necessary condition is not verified, then the series does not converge; |x| = 1 ⇔ { x = 1, +∞Σk=1 1/k = +∞; x = -1, +∞Σk=1 (-1)k/k < +∞}; +∞Σk=1 xk/k = {absolutely converges if |x| < 1; converges if x = -1; does not converge if |x| > 1 and x = 1}

+∞Σk=1 (3k-3k)/(5k-5k) = - +∞Σk=1 (3k-3k)/(5k-5k); (3k-3k)/(5k-5k) ~ 3k/5k = (3/5)k; +∞Σk=1 (3/5)k < +∞ ⇒ +∞Σk=1 (3k-3k)/(5k-5k) converges

+∞Σk=1 (√k+1-√k)/k; √k+1-√k > 0, it is a positive term series; limk→∞ (√k+1-√k)/k = 0, the necessary condition is verified; ((√k+1-√k)/k)((√k+1+√k)/(√k+1+√k)) = (k+1-k)/(k(√k+1+√k)) = 1/(k(√k+1+√k)) ~ 1/2k√k = 1/2k3/2, considering that √k+1+√k = √k(√(k+1)/k+1); +∞Σk=1 1/k3/2 < +∞ ⇒ +∞Σk=1 (√k+1-√k)/k < +∞

+∞Σk=1 (k!+2)/(k+2)!; (k!+2)/(k+2)! ~ k!/(k+2)! = k!/k!(k+1)(k+2) = 1/(k+1)(k+2) ~ 1/k2; +∞Σk=1 1/k2 < +∞ ⇒ +∞Σk=1 (k!+2)/(k+2)! < +∞

+∞Σk=1 (-1)k/log(k+1); according to the Leibniz criterion, +∞Σk=1 (-1)k ak, ak = 1/log(k+1), if 1) limk→∞ ak = 0 and 2) ak > ak+1 ∀ k ⇒ +∞Σk=1 (-1)k ak converges; 1) limk→∞ 1/log(k+1) = 0 and 2) ak > ak+1 ⇔ 1/log(k+1) > 1/log(k+2) ⇔ log(k+2) > log(k+1), then +∞Σk=1 (-1)k/log(k+1) converges

{ak}k∈ℕ | limk→∞ ak = 1/2, bk = (ak)k, then, a) +∞Σk=0 bk is irregular, b) +∞Σk=0 bk diverges, c) +∞Σk=0 bk converges, d) none of the previous answers; ak = 1/2 ∀ k ∈ ℕ, bk = (1/2)k+∞Σk=0 (1/2)k < +∞; limk→∞ ak = 1/2, then ak ≤ 2/3 ∀ k ∈ ℕ, 0 ≤ bk = (ak)k ≤ (2/3)k, the series bk is upper bounded by a converging geometric series; +∞Σk=0 (2/3)k converges ⇒ according to the comparison theorem +∞Σk=0 bk < +∞