EXERCISES - SERIES

^{+∞}Σ_{k=0} k/(k^{2}+1); necessary condition lim_{k→∞} a_{k} = lim_{k→∞} k/(k^{2}+1) = 0; it is a positive term series, a_{k} = k/(k^{2}+1) ≥ 0 ∀ k ∈ ℕ; lim_{k→∞} a_{k}/b_{k} = l ≠ 0, then ^{+∞}Σ_{k=0} a_{k} < +∞ ⇔ ^{+∞}Σ_{k=0} b_{k} < +∞, the two series have the same behavior; k/(k^{2}+1) ~ k/k^{2} = 1/k ⇔ lim_{k→∞} (k/(k^{2}+1))/(1/k) = 1; ^{+∞}Σ_{k=1} 1/k = +∞ ⇒ ^{+∞}Σ_{k=1} k/(k^{2}+1) = +∞

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

^{+∞}Σ_{k=1} ln(k)/k^{3}; lim_{k→∞} ln(k)/k^{3} = 0; ln(k) ≥ 0 ∀ k ∈ ℕ, k ≥ 1, it is a positive term series; ln(k)/k^{3} ~ ?/k^{3}; according to the second part of the asymptotic comparison criterion, if lim_{k→∞} a_{k}/b_{k} = 0, then ^{+∞}Σ_{k=0} b_{k} < +∞ ⇒ ^{+∞}Σ_{k=0} a_{k} < +∞; lim_{k→∞} (ln(k)/k^{3})/(1/k^{2}) = lim_{k→∞} (ln(k)/k^{3})k^{2} = 0; ^{+∞}Σ_{k=1} 1/k^{2} < +∞ ⇒ ^{+∞}Σ_{k=1} ln(k)/k^{3} < +∞; ^{+∞}Σ_{k=1} 1/k^{α} = {< +∞, α > 1; = +∞, α ≤ 1}

^{+∞}Σ_{k=1} ln(k)/k^{2} < +∞; lim_{k→∞} ln(k)/k^{2} = 0; ln(k) ≥ 0 ∀ k ∈ ℕ, k ≥ 1, it is a positive term series; ln(k)/k^{2} ~ ?/k^{2}; according to the second part of the asymptotic comparison criterion, if lim_{k→∞} a_{k}/b_{k} = 0, then ^{+∞}Σ_{k=0} b_{k} < +∞ ⇒ ^{+∞}Σ_{k=0} a_{k} < +∞; lim_{k→∞} (ln(k)/k^{2})/(1/k^{3/2}) = lim_{k→∞} ln(k)/k^{1/2} = 0; ^{+∞}Σ_{k=1} 1/k^{3/2} < +∞ ⇒ ^{+∞}Σ_{k=1} ln(k)/k^{2} < +∞; lim_{k→∞} ln(k)/k^{α} = 0 ∀ α > 0; ^{+∞}Σ_{k=1} 1/k^{α} = {< +∞, α > 1; = +∞, α ≤ 1}

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

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

^{+∞}Σ_{k=1} (3k-3^{k})/(5^{k}-5k) = - ^{+∞}Σ_{k=1} (3^{k}-3k)/(5^{k}-5k); (3^{k}-3k)/(5^{k}-5k) ~ 3^{k}/5^{k} = (3/5)^{k}; ^{+∞}Σ_{k=1} (3/5)^{k} < +∞ ⇒ ^{+∞}Σ_{k=1} (3k-3^{k})/(5^{k}-5k) converges

^{+∞}Σ_{k=1} (√k+1-√k)/k; √k+1-√k > 0, it is a positive term series; lim_{k→∞} (√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 = 1/2k^{3/2}, considering that √k+1+√k = √k(√(k+1)/k+1); ^{+∞}Σ_{k=1} 1/k^{3/2} < +∞ ⇒ ^{+∞}Σ_{k=1} (√k+1-√k)/k < +∞

^{+∞}Σ_{k=1} (k!+2)/(k+2)!; (k!+2)/(k+2)! ~ k!/(k+2)! = k!/k!(k+1)(k+2) = 1/(k+1)(k+2) ~ 1/k^{2}; ^{+∞}Σ_{k=1} 1/k^{2} < +∞ ⇒ ^{+∞}Σ_{k=1} (k!+2)/(k+2)! < +∞

^{+∞}Σ_{k=1} (-1)^{k}/log(k+1); according to the Leibniz criterion, ^{+∞}Σ_{k=1} (-1)^{k} a_{k}, a_{k} = 1/log(k+1), if 1) lim_{k→∞} a_{k} = 0 and 2) a_{k} > a_{k+1} ∀ k ⇒ ^{+∞}Σ_{k=1} (-1)^{k} a_{k} converges; 1) lim_{k→∞} 1/log(k+1) = 0 and 2) a_{k} > a_{k+1} ⇔ 1/log(k+1) > 1/log(k+2) ⇔ log(k+2) > log(k+1), then ^{+∞}Σ_{k=1} (-1)^{k}/log(k+1) converges

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