A more difficult assertion about series

Task number: 2937

Show that if the series \(\displaystyle \sum_{n=1}^{\infty}a_n \) with positive terms diverges and \(\displaystyle s_n=\sum_{k=1}^{n}a_k \) is the sequence of partial sums, then the series \(\displaystyle \sum_{n=1}^{\infty}\frac{a_n}{s_n} \) diverges and the series \(\displaystyle \sum_{n=1}^{\infty}\frac{a_n}{s_n^2} \) converges.

  • Hint

    For divergence use the Bolzano-Cauchy criterion.

    For convergence, write \(a_n=s_n-s_{n-1}\).

  • Resolution

    Let \(\displaystyle b_n=\sum_{k=1}^n \frac{a_n}{s_n}\).

    Then \(b_m-b_n= \frac{a_{n+1}}{s_{n+1}}+\frac{a_{n+2}}{s_{n+2}}+…+\frac{a_{m}}{s_{m}}\ge \frac{a_{n+1}}{s_{m}}+\frac{a_{n+2}}{s_{m}}+…+\frac{a_{m}}{s_{m}}= \frac{s_m-s_n}{s_m}=1-\frac{s_n}{s_m} \)

    But because \(\displaystyle \lim_{n\to\infty} s_n=\infty\), for every \(n\) we can choose a sufficiently large \(m\) such that \(s_m\ge 2s_n\).

    But then \(b_m-b_n\ge\frac12\), so the series \(\displaystyle \sum_{n=1}^{\infty}\frac{a_n}{s_n} \) diverges.

    If we omit the first term of the series, we obtain the estimate

    \(\displaystyle \sum_{n=2}^{\infty}\frac{a_n}{s_n^2} = \sum_{n=2}^{\infty}\frac{s_n-s_{n-1}}{s_n^2} \le \sum_{n=2}^{\infty}\frac{s_n-s_{n-1}}{s_ns_{n-1}} = \sum_{n=2}^{\infty}\left(\frac1{s_{n-1}}-\frac1{s_{n}}\right)=\frac1{a_1}\)

    So the series \(\displaystyle \sum_{n=1}^{\infty}\frac{a_n}{s_n^2} \) converges by the comparison test.

Difficulty level: Hard task
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