Exercise 2.7.13

Question

[Abel's Test]
  Abel's Test for convergence states that if the series $\sum_{k=1}^{\infty} x_{k}$ converges, and if $\left(y_{k}\right)$ is a sequence satisfying
  $$
  y_{1} \geq y_{2} \geq y_{3} \geq \cdots \geq 0
  $$
  then the series $\sum_{k=1}^{\infty} x_{k} y_{k}$ converges.
  \begin{enumerate}[label=(\alph*)] 
  \item Use Exercise $2.7 .12$ to show that
    $$
    \sum_{k=1}^{n} x_{k} y_{k}=s_{n} y_{n+1}+\sum_{k=1}^{n} s_{k}\left(y_{k}-y_{k+1}\right)
    $$
    where $s_{n}=x_{1}+x_{2}+\cdots+x_{n}$.
  \item Use the Comparison Test to argue that $\sum_{k=1}^{\infty} s_{k}\left(y_{k}-y_{k+1}\right)$ converges absolutely, and show how this leads directly to a proof of Abel's Test.
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item Exercise 2.7.12 combined with $s_0=0$ gives
    $$
    \sum_{k=1}^n x_ky_k = s_ny_{n+1} + \sum_{k=1}^n s_k(y_k - y_{k+1})
    $$
    as desired.
  \item $s_ny_{n+1}$ clearly converges since $y_{n+1}$ is ``eventually constant'', so we must only show the right hand side converges.

We will show absolute convergence, note $y_k - y_{k+1} \ge 0$ and so
    $$
    \sum_{k=1}^n |s_k|(y_k - y_{k+1}) \ge 0
    $$
    Bounding $|s_n| \le M$ gives
    $$
    \sum_{k=1}^n |s_k|(y_k - y_{k+1}) \le M \sum_{k=1}^n (y_k-y_{k+1})
    $$
    Since $\sum_{k=1}^n (y_k - y_{k+1}) = y_1 - y_{n+1}$ is telescoping we can write
    $$
    \sum_{k=1}^n |s_k|(y_k - y_{k+1}) \le M(y_1 - y_{n+1}) \le My_1
    $$
    Implying $\sum_{k=1}^\infty |s_k|(y_k - y_{k+1})$ converges since it is bounded and increasing.
    And since the series converges absolutely so does the original $\sum_{k=1}^\infty s_k(y_k - y_{k+1})$.

extbf{Summary:} Bound $|s_k| \le M$ and use the fact that $\sum(y_k - y_{k+1})$ is telescoping.
   \end{enumerate}

Exercise 2.7.13

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Exercise 2.7.13

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