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I'm reading through Spivak Ch.22 (Infinite Sequences) right now. He mentioned in the written portion that it's often not a trivial matter to determine the boundedness of sequences. With that in mind, he gave us a sequence to chew on before we learn more about boundedness. That sequence is:

$$1, 1+\frac{1}{2}, 1+\frac{1}{2}+\frac{1}{3}, 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}, . . .$$

I know that a sequence is bounded above if there is a number $M$ such that $a_n\leq M$ for all $n$. Any hints here?

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    $\begingroup$ It is good that you are attending to the definition of boundedness. So, can you write down a suitable $M$ with $a_n\le M$ for all $n$? $\endgroup$ – Angina Seng Jan 10 '19 at 5:08
  • $\begingroup$ @LordSharktheUnknown I'm not too sure what that $M$ would be. Can you give a hint? $\endgroup$ – psa Jan 10 '19 at 5:14
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    $\begingroup$ Those aren't subsequences. In any case, are you sure you have transcribed Spivak's sequence correctly? $\endgroup$ – Angina Seng Jan 10 '19 at 5:17
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    $\begingroup$ It is just divergence of harmonic series. You can search the net for 'divergence of harmonic series'. $\endgroup$ – Kavi Rama Murthy Jan 10 '19 at 5:23
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    $\begingroup$ A classic estimate: for each $n$, consider estimating $a_{2^n}$. $\endgroup$ – xbh Jan 10 '19 at 5:26
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This is a classic example of how our intuition can be wrong concerning infinities. As @Kavi said in the comments, this is the harmonic series and indeed diverges. I believe that Spivak provided this as an example to “chew on” in order to show how non trivial boundedness can be, as on first glance many people believe this sequence should be bounded.

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    $\begingroup$ Indeed! Thank you for that input. $\endgroup$ – psa Jan 10 '19 at 5:57
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    $\begingroup$ This is not the harmonic series, it is the sequence of partial sums of the harmonic series. A sequence and a series are two different things. Other important sequences linked to a series are the sequence of the terms ($\{1/n\}$ in this case) and the sequence of partial rests. $\endgroup$ – A. Arredondo Jan 10 '19 at 9:16
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Indeed, you can notice that

$$a_1=1,a_4>2(=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{4}), a_8>\frac{5}{2}(=a_4+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}),a_{16}>3(=a_{8}+\frac{1}{16}+\frac{1}{16}+..8times)$$, so the general pattern is $a_{2^{2n}}>n+1$, so given an upper bound M, you can find a natural number n s.t. $n\geq M$ by archimedian property and so $a_{2^{2n}}>n+1>M$ and hence it can't be bounded.

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