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Prove some theorems on convergence of series

by RoRi

Consider two sequences \{ a_n \} and \{ b_n \} whose terms are always positive after some integer N (i.e., a_n > 0 and b_n > 0 for all n \geq N). Then define

    \[ c_n = b_n - \frac{b_{n+1}a_{n+1}}{a_n}. \]

Prove the following two statements.

  1. If there exists a constant r with 0 < r \leq c_n for all n \geq N, then \sum a_n converges.
  2. If the series \sum \frac{1}{b_n} diverges and if c_n \leq 0 for all n \geq 0 then the series \sum a_n diverges.
  1. Proof. First, we solve the given relation for b_n (starting with the term N so that all terms are positive, hence, nonzero),

        \begin{align*} && c_N &= b_N - \frac{b_{N+1} a_{N+1}}{a_N} \\[9pt] \implies && b_N &= c_N + \left( \frac{a_{N+1}}{a_N} \right) b_{N+1} \\[9pt] \implies && b_N &= c_N + \left( \frac{a_{N+1}}{a_N} \right) c_{N+1} + \left( \frac{a_{N+2}}{a_N} \right) b_{N+2}\\[9pt] \implies && b_N &= c_N + \left( \frac{a_{N+1}}{a_N} \right) c_{N+1} + \left( \frac{a_{N+2}}{a_N} \right) c_{N+2} + \left( \frac{a_{N+3}}{a_N} \right)b_{N+3} \\[9pt] &&& \vdots \\[9pt] \implies && b_N &= c_N + \left( \frac{a_{N+1}}{a_N} \right) c_{N+1} + \cdots + \left( \frac{a_{N+n}}{a_N} c_{N+n} + \left( \frac{a_{N+n+1}}{a_N} \right) b_{N+n+1}. \end{align*}

    Then, since 0 < c_N \leq r for all n \geq N, we have

        \begin{align*} \implies && b_N &\geq r \left( 1 + \frac{a_{N+1}}{a_N} + \cdots + \frac{a_{N+n}}{a_N} \right) + \frac{a_{N+n+1}}{a_N} b_{N+n+1} \\[9pt] \implies && b_N &> r \left( 1 + \frac{a_{N+1}}{a_N} + \cdots + \frac{a_{N+n}}{a_N} \right) \\[9pt] \implies && \frac{b_N a_N}{r} &> a_N + a_{N+1} + \cdots + a_{N+n} = \sum_{k=N}^n a_k. \end{align*}

    Therefore, the partial sums of \sum a_n are bounded (since \frac{a_N b_N}{r} is a constant); hence, \sum a_n converges. \qquad \blacksquare

  2. Proof. We are given c_n \leq 0 which implies

        \[ b_n - \frac{b_{n+1}a_{n+1}}{a_n} \leq 0 \quad \implies \quad b_n \leq \frac{b_{n+1}a_{n+1}}{a_n}. \]

    This then implies

        \begin{align*} && b_n a_n &\leq b_{n+1} a_{n+1} \\[9pt] \implies && \frac{a_n}{\frac{1}{b_n}} &\leq \frac{a_{n+1}}{\frac{1}{b_{n+1}}} & \text{for all } n \geq N \\[9pt] \implies && \frac{a_N}{\frac{1}{b_N}} &\leq \frac{a_n}{\frac{1}{b_n}} \text{for all } n \geq N\\[9pt] \implies && \left( \frac{1}{b_N} \right) \left( \frac{a_N}{\frac{1}{b_N}} \right) &\leq a_n &\text{for all } n \geq N \\[9pt] \implies && \sum a_N \geq \left( \frac{a_N}{\frac{1}{b_N}} \right) \sum \frac{1}{b_n}. \end{align*}

    Since \frac{a_N}{\frac{1}{b_N}} is a constant this implies that if \sum \frac{1}{b_n} diverges so does \sum a_n. \qquad \blacksquare

4 comments

  1. Anonymous says:

    Here is my proof for part a).

    we know from the problem statement

        \[ b_n - b_{n+1} \frac{a_{n+1}}{a_n} \ge r > 0 \]

    multiplying by a_n and dividing by r we get

        \[ \frac{1}{r}\left(b_na_n - b_{n+1} a_{n+1} \right) \ge a_n > 0 \]

    summing up such inequalities for k=1 to k=n we have

        \[ \frac{1}{r}\sum_{k=1}^{n}\left(b_ka_k - b_{k+1} a_{k+1} \right) \ge \ sum_{k=1}^{n}a_k > 0 \]

    the first sum collapses by the telescopic property

        \[ \frac{1}{r} \left( b_1a_1 - b_{n+1}a_{n+1} \right) \ge \sum_{k=1}^{n} a_k > 0 \]

    since a_n >0 \land b_n > 0 we have

        \[ \frac{1}{r} b_1a_1 > \sum_{k=1}^{n}a_k > 0 \]

    the sequence of partial sums of a_n is thus bounded. And since it’s increasing too (a_n >0), then it must converge by theorem 10.1.

    • Mohammad Azad says:

      I solved it the same way! I will add my proof of (b) :
      $$b_n－\frac {b_{n+1}a_{n＋1}}{a_n}≤0$$ $$a_nb_n≤b_{n+1}a_{n＋1}$$
      $$0≤a_Nb_N≤a_nb_n$$
      $$0≤\frac {a_Nb_N}{b_n}≤a_n$$

  2. Anonymous says:

    Just noticed a couple of typos. See suggestions below.

    In the proof for part a:
    \begin{frame}
    Then, since 0 < r \leq c_N for all n \geq N, we have
    \end{frame}

    In the proof for part b:
    \begin{frame}

        \begin{align*} \implies && \left( \frac{1}{b_n} \right) \left( \frac{a_N}{\frac{1}{b_N}} \right) &\leq a_n &\text{for all } n \geq N \\[9pt] \implies && \sum a_n \geq \left( \frac{a_N}{\frac{1}{b_N}} \right) \sum \frac{1}{b_n}. \end{align*}

    \end{frame}

  3. Anonymous says:

    In the proof for part a, should the sentence “Then, since 0 < c_N \leq r for all n \geq N, we have" be "Then, since 0 < c_N \leq r for all n \geq N, we have" instead?

    In the proof for part b,

        \begin{align*} \implies && \left( \frac{1}{b_n} \right) \left( \frac{a_N}{\frac{1}{b_N}} \right) &\leq a_n &\text{for all } n \geq N \\[9pt] \implies && \sum a_n \geq \left( \frac{a_N}{\frac{1}{b_N}} \right) \sum \frac{1}{b_n}. \end{align*}

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