Exercise 2.7.14

Question

[Dirichlet's Test]
  Dirichlet's Test for convergence states that if the partial sums of $\sum_{k=1}^{\infty} x_{k}$ are bounded (but not necessarily convergent), and if $\left(y_{k}\right)$ is a sequence satisfying $y_{1} \geq y_{2} \geq y_{3} \geq \cdots \geq 0$ with $\lim y_{k}=0$, then the series $\sum_{k=1}^{\infty} x_{k} y_{k}$ converges.
  \begin{enumerate}[label=(\alph*)] 
  \item Point out how the hypothesis of Dirichlet's Test differs from that of Abel's Test in Exercise 2.7.13, but show that essentially the same strategy can be used to provide a proof.
  \item Show how the Alternating Series Test (Theorem 2.7.7) can be derived as a special case of Dirichlet's Test.
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item
    Abel's test gets it's convergence from $\sum x_n$ converging, while Dirichlet's test gets its convergence from $(y_n) 	o 0$.
    Expanding on that, the proof that the $\sum_{k=1}^n s_k(y_k-y_{k-1})$ term converges is the same in Abel and Dirichlet, but the proof that $s_ny_{n+1}$ differs depending on if we get our convergence from $(y_n)	o0$ and $(s_n)$ bounded, or $(s_n) 	o s$ and $(y_n)$ bounded.

The proof is almost identical to Abel's test, bound $|s_n|<M$ and use the triangle inequality on the right hand side to get (note $y_k - y_{k+1} > 0$ because decreasing)
    $$
    \sum_{k=1}^n |s_k|(y_k - y_{k+1}) \le M \sum_{k=1}^n (y_k - y_{k+1}) = M(y_1 - y_{n+1}) \le My_1
    $$
    Thus $\sum s_k(y_k - y_{k+1})$ is bounded and increasing, so it converges by MCT.
    To see the $s_ny_{n+1}$ term converges, simply note that $(y_n) 	o 0$ and $|s_n|<M$.

Thus the original series converges, and furthermore
    $$
    \sum_{k=1}^\infty x_ky_k  = \sum_{k=1}^\infty s_k(y_k - y_{k+1})
    $$
    Because $(s_ny_{n+1}) 	o 0$.
  \item
    Let $a_n \ge 0$ with $a_1 \ge a_2 \ge \dots \ge 0$ and $\lim a_n = 0$.
    The series $\sum (-1)^n$ is bounded, so Dirichlet's test implies $\sum (-1)^n a_n$ converges.
   \end{enumerate}

Exercise 2.7.14

Answers

Comments

Exercise 2.7.14

Answers

Comments

Add answer