If $u_n>0$ and $\sum u_n$ diverges and $s_n=u_1+u_2+\dots+u_n$ then is the claim that "we can choose $p\in \mathbb{N}$ such that $s_{n+p}>2s_n$" is justified?

I know the following

  1. $\sum u_n$ diverges iff sequence of partial sum $\{s_n\}$ diverges

  2. sequence $\{s_n\}$ is monotone increasing

But how to show that $s_{n+p}>2s_n$ is justified?

In the link Is my proof ok? If $\sum u_n$ diverges then $\sum \frac {u_n} {u_1 + u_2 + \dots + u_n}$ also diverges

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Only want to get the result $s_{n+p}>2s_n$. Please help

  • $\begingroup$ Since your method does not work I have provided a different argument to show that $\sum \frac {u_n} {s_n}$ is divergent. See the answer below. $\endgroup$ – Kabo Murphy May 30 '18 at 8:24

If $u_n = 1$, this sequence verifies your hypotheses. However, $s_n = n$, so

$$ s_{n+p} > 2s_n \iff n+p > 2n \iff p >n $$

which of course for a fixed $p$ it cannot hold for all $n \in \mathbb{N}$.

  • $\begingroup$ Do you mean that $s_{n+p} > 2s_n$ is not true $\endgroup$ – user1942348 May 30 '18 at 11:17
  • $\begingroup$ @user1942348 For this example, it will be false every time that $ n > p $. $\endgroup$ – Guido A. May 30 '18 at 18:53

Suppose $u_n >0$ for all $n$ and $\sum u_n =\infty$. If $\sum \frac {u_k} {s_k}$ is convergent then $\sum_{k=n}^{m}\frac {u_k} {s_k} \to 0$ as $n,m\to \infty$. We prove that this is not the case. We need the following inequality: $1-x_1 x_2...x_j \leq \sum_{i=1}^{j} (1-x_i)$ if each $x_i \in [0,1]$. This inequality can be proved by induction. From this we get $\sum_n^{m}\frac {u_k} {s_k}=\sum_n^{m}\frac {s_k-s_{k-1}} {s_k} \geq 1-\prod_{k=n}^{m}\frac {s_{k-1}} {s_k} =1-\frac {s_{n-1}} {s_m} \to 1$ as $m \to \infty$. This completes the proof.

  • 1
    $\begingroup$ I request you for the proof of $s_{n+p}>2s_n$ $\endgroup$ – user1942348 May 30 '18 at 10:24

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