9Math I - Exercise Sheet 9: Series Convergence and the Basel Problem and Convergence Criteria for Convergence

Exercise 1: Convergence of Series

• Subtask 1a: Root Test Analysis

  • The series is defined as: $\sum_{k=1}^{\infty} \left( \frac{k}{k^2 + 1} \right)^k$

  • Application of the Root Criterion (Wurzelkriterium): Since the terms contain powers to the degree $k$, we evaluate the expression $\sqrt[k]{|a_k|}$.

  • Calculation: $\sqrt[k]{\left| \left( \frac{k}{k^2 + 1} \right)^k \right|} = \frac{k}{k^2 + 1}$.

  • Determining the limit: $\lim_{k \to \infty} \sqrt[k]{|a_k|} = \lim_{k \to \infty} \frac{k}{k^2 + 1} = \lim_{k \to \infty} \frac{\frac{1}{k}}{1 + \frac{1}{k^2}} = 0$.

  • The solution notes that Bernoulli's rule and de l'Hospital's rule were applied in the steps prior to the limit transition.

  • Conclusion: Since the limit $0 < 1$, the Root Criterion proves the series converges.

• Subtask 1b: Quotient Test Analysis

  • The series is defined as: $\sum_{k=0}^{\infty} \frac{2^k}{k!}$.

  • Application of the Quotient Criterion (Quotientenkriterium): We analyze the ratio $\left| \frac{a_{k+1}}{a_k} \right|$.

  • Calculation: $\frac{a_{k+1}}{a_k} = \frac{2^{k+1}}{(k+1)!} \times \frac{k!}{2^k} = \frac{2 \times 2^k}{(k+1) \times k!} \times \frac{k!}{2^k} = \frac{2}{k+1}$.

  • Determining the limit: $\lim_{k \to \infty} \frac{a_{k+1}}{a_k} = \lim_{k \to \infty} \frac{2}{k+1} = 0$.

  • Conclusion: Since the limit $0 < 1$, the series converges.

• Subtask 1c: Quotient Test Analysis

  • The series is defined as: $\sum_{k=0}^{\infty} \frac{2^k}{e^k}$.

  • Application of the Quotient Criterion: $\frac{a_{k+1}}{a_k} = \frac{2^{k+1}}{e^{k+1}} \times \frac{e^k}{2^k} = \frac{2 \times 2^k}{e \times e^k} \times \frac{e^k}{2^k} = \frac{2}{e}$.

  • Determining the limit: $\lim_{k \to \infty} \frac{a_{k+1}}{a_k} = \frac{2}{e} \approx 0.736$.

  • Conclusion: Since $\frac{2}{e} < 1$, the Quotient Criterion proves the series converges.

Exercise 2: Applications of Series Convergence and the Necessary Criterion

• Subtask 2a: Convergence Condition for Parameter $\alpha$

  • The series is defined as: $\sum_{k=0}^{\infty} e^{-\alpha k}$.

  • Using the Root Criterion: $\sqrt[k]{|a_k|} = \sqrt[k]{(e^{-\alpha})^k} = e^{-\alpha}$.

  • Limit evaluation: $\lim_{k \to \infty} \sqrt[k]{|a_k|} = \lim_{k \to \infty} e^{-\alpha} = e^{-\alpha}$.

  • Condition for convergence: The series converges if and only if $e^{-\alpha} < 1$.

  • Solving for $\alpha$: $e^{-\alpha} < 1 \iff -\alpha < 0 \iff \alpha > 0$.

  • Result: The series converges for all positive real numbers $\alpha \in \mathbb{R}$.

• Subtask 2b: Calculation of Series Value

  • The series $\sum_{k=0}^{\infty} e^{-\alpha k}$ is treated as a geometric series $\sum_{k=0}^{\infty} q^k$ where $q = e^{-\alpha}$.

  • Standard geometric series formula: $\sum_{k=0}^{\infty} q^k = \frac{1}{1 - q}$ provided $|q| < 1$.

  • Final Result: Substituting $q = e^{-\alpha}$ gives the value $\frac{1}{1 - e^{-\alpha}}$.

• Subtask 2c: Necessary Criterion and Null Sequences

  • The series is defined as: $\sum_{k=0}^{\infty} k e^{-k}$.

  • Quotient Criterion Application: $\frac{a_{k+1}}{a_k} = \frac{(k+1) e^{-(k+1)}}{k e^{-k}} = \frac{k+1}{k} \times \frac{e^{-k} \times e^{-1}}{e^{-k}} = \left( 1 + \frac{1}{k} \right) \times \frac{1}{e}$.

  • Limit evaluation: $\lim_{k \to \infty} \frac{a_{k+1}}{a_k} = \lim_{k \to \infty} \left( 1 + \frac{1}{k} \right) \times \frac{1}{e} = 1 \times \frac{1}{e} = \frac{1}{e} < 1$.

  • Implication for the sequence: Because the series $\sum k e^{-k}$ converges, the sequence of its terms $(k e^{-k})$ must be a null sequence (Nullfolge) according to the necessary criterion for series convergence (notwendiges Kriterium).

  • Conclusion: $\lim_{k \to \infty} k e^{-k} = 0$.

Exercise 3: Convergence Radius of Power Series

• Subtask 3a: Quotient Criterion for Radius

  • Series: $\sum_{k=1}^{\infty} \frac{k+1}{k} x^k$.

  • Calculation: $\left| \frac{a_{k+1}}{a_k} \right| = \left| \frac{(k+2)/(k+1) \times x^{k+1}}{(k+1)/k \times x^k} \right| = \left| \frac{k+2}{k+1} \times \frac{k}{k+1} \right| \times |x| = \frac{k^2 + 2k}{k^2 + 2k + 1} \times |x|$.

  • Limit: $\lim_{k \to \infty} \frac{k^2 + 2k}{k^2 + 2k + 1} |x| = 1 \times |x| = |x|$.

  • Radius: Convergence occurs for $|x| < 1$, so the convergence radius $R = 1$.

• Subtask 3b: Cosine Function Taylor Expansion

  • Series: $\sum_{k=0}^{\infty} \frac{(-1)^k x^{2k}}{(2k)!}$.

  • Calculation: $\left| \frac{a_{k+1}}{a_k} \right| = \left| \frac{x^{2(k+1)}}{(2(k+1))!} \times \frac{(2k)!}{x^{2k}} \right| = \frac{x^{2k} \times x^2}{(2k+2) \times (2k+1) \times (2k)!} \times \frac{(2k)!}{x^{2k}} = \frac{x^2}{(2k+2)(2k+1)}$.

  • Limit: $\lim_{k \to \infty} \frac{x^2}{4k^2 + 6k + 2} = 0$.

  • Radius: Since the limit is $0 < 1$ for all $x \in \mathbb{R}$, the convergence radius $R = \infty$. This series represents the Taylor expansion of the cosine function.

• Subtask 3c: Exponential Function Expansion

  • Series: $e^x = \sum_{k=0}^{\infty} \frac{x^k}{k!}$.

  • Calculation: $\left| \frac{a_{k+1}}{a_k} \right| = \left| \frac{x^{k+1}}{(k+1)!} \times \frac{k!}{x^k} \right| = \frac{|x|}{k+1}$.

  • Limit: $\lim_{k \to \infty} \frac{|x|}{k+1} = 0$.

  • Radius: The limit is 0 regardless of the choice of $|x|$, thus the series converges for all $x \in \mathbb{R}$ and $R = \infty$.

• Subtask 3d: Shifted Power Series

  • Series: $\ln(x) = \sum_{k=1}^{\infty} (-1)^{k+1} (x - 1)^k$.

  • Root Criterion Application: $\sqrt[k]{|a_k|} = \sqrt[k]{|(x-1)^k|} = |x-1|$.

  • Limit: $\lim_{k \to \infty} |x-1| = |x-1|$.

  • Convergence: The series converges if $|x-1| < 1$, which implies $0 < x < 2$.

  • Conclusion: The series was developed around the point $x_0 = 1$, and the convergence radius is $R = 1$.

• Subtask 3e: Hyperbolic Tangent Series Variation

  • Expansion: $\tanh(x) = -1 + 2 \sum_{k=0}^{\infty} (-1)^k q^{2k}$ where $q = e^x$.

  • Root Criterion on the sum portion: $\sqrt[k]{|(-1)^k q^{2k}|} = \sqrt[k]{|q^{2k}|} = q^2$.

  • Limit: $\lim_{k \to \infty} q^2 = q^2$.

  • Note: Since $q = e^x$ is always positive for any $x \in \mathbb{R}$, absolute value bars can be omitted.

  • Convergence Condition: $q^2 < 1 \iff 0 < q < 1$.

  • Substituting back for $x$: $0 < e^x < 1 \iff \ln(0) < x < \ln(1) \iff -\infty < x < 0$.

  • Result: This variant of the tanh-series converges for $x \in (-\infty, 0)$.

Exercise 4: The Basel Problem (Additional Task)

• Subtask 4a: Failure of Ratio and Root Tests

  • The task is to evaluate the series $\sum_{k=1}^{\infty} \frac{1}{k^2}$.

  • Applying the Quotient Criterion: $\frac{a_{k+1}}{a_k} = \frac{1/(k+1)^2}{1/k^2} = \frac{k^2}{k^2 + 2k + 1}$.

  • Limit determination: $\lim_{k \to \infty} \frac{k^2}{k^2 + 2k + 1} = \lim_{k \to \infty} \frac{1}{1 + 2/k + 1/k^2} = 1$.

  • Finding: Both the Root and Quotient criteria result in a limit of 1. Therefore, no statement regarding convergence can be made using these methods.

• Subtask 4b: Partial Fraction Decomposition Proof

  • Objective: Show that $\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}$.

  • Verification: $\frac{1}{k} - \frac{1}{k+1} = \frac{(k+1) - k}{k(k+1)} = \frac{1}{k(k+1)}$.

  • The identity is verified.

• Subtask 4c: Evaluation via Telescoping Series and Majorant Criterion

  • Calculating the series value for $\sum_{k=1}^{\infty} \frac{1}{k(k+1)}$ using the result from 4b:

  • The sequence of partial sums $s_n = \sum_{k=1}^n \left( \frac{1}{k} - \frac{1}{k+1} \right)$.

  • Telescoping expansion: $(1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + \dots + (1/n - 1/(n+1)) = 1 - 1/(n+1)$.

  • Determining the series sum: $\lim_{n \to \infty} (1 - 1/(n+1)) = 1$. Thus, $\sum_{k=1}^{\infty} \frac{1}{k(k+1)} = 1$.

  • Majorant Comparison: We utilize the relation $\frac{1}{k+1} < \frac{1}{k}$.

  • Inequality: $\sum \frac{1}{(k+1)(k+1)} < \sum \frac{1}{k(k+1)}$.

  • This establishes that $\sum \frac{1}{k(k+1)}$ is a convergent majorant for $\sum \frac{1}{(k+1)^2}$.

  • Conclusion: By the Majorant Criterion, since the larger series converges, $\sum \frac{1}{(k+1)^2}$ and consequently $\sum \frac{1}{k^2}$ must converge.