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We are given a sequence $(a_n)$ such that $0\lt a_n\leq a_{2n}+a_{2n+1}$ for all natural numbers $n$. Show that $\sum_{n=1}^\infty a_n$ diverges.

I attempted the question as follows. Consider, say, $$\begin{align} S_7 &= a_1+(a_2+a_3)+(a_4+a_5)+(a_6+a_7)\\ &\geq a_1+a_1+(a_2+a_3)\\ &\geq a_1+a_1+a_1 \end{align}$$

But I have no idea how to generalise this to show that the sequence diverges.

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    $\begingroup$ You're not new here -- you must know the guidelines... what are your thoughts on the question? $\endgroup$
    – Clement C.
    Commented Sep 13, 2017 at 12:34
  • $\begingroup$ Hint: Try a proof by contradiction. Suppose the series $\sum_{n=1}^\infty a_n$ converges to $\ell$. Now, try to arrive at a contradiction using the mentioned inequality. $\endgroup$
    – Prasun Biswas
    Commented Sep 13, 2017 at 12:40
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    $\begingroup$ Hint #2: Notice that if the series converges, we have $\ell\leq\ell -a_1$ and furthermore we know that $a_1\gt 0$. Can you take it from here? $\endgroup$
    – Prasun Biswas
    Commented Sep 13, 2017 at 12:44
  • $\begingroup$ @PrasunBiswas - Sorry but how did you arrive at $l\leq l-a_1$? $\endgroup$
    – user443782
    Commented Sep 13, 2017 at 12:49
  • $\begingroup$ @KaranKaran I guess, by summing the inequality from $n=1$ to $\infty$. $\endgroup$
    – Clement C.
    Commented Sep 13, 2017 at 12:51

1 Answer 1

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What you found can indeed be generalized. Here is a hint:

Take any natural number $n$.

Then $\displaystyle \sum_{k=1}^{2^{n+1}-1} a_k = a_1 + \sum_{k=1}^{2^n-1} ( a_{2k}+a_{2k+1} ) \ge a_1 + \sum_{k=1}^{2^n-1} a_k$.

Thus $\displaystyle \sum_{k=1}^{2^n-1} a_k - \sum_{k=1}^{2^0-1} a_k \ge n·a_1$, by induction.

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