# Is the sequence of sums of inverse of natural numbers bounded? [duplicate]

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?

## marked as duplicate by Arnaud D., Martin R, user159517, Saad, Matthew TowersJan 10 at 16:24

• 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 $n$? – Lord Shark the Unknown Jan 10 at 5:08
• @LordSharktheUnknown I'm not too sure what that $M$ would be. Can you give a hint? – kyle campbell Jan 10 at 5:14
• Those aren't subsequences. In any case, are you sure you have transcribed Spivak's sequence correctly? – Lord Shark the Unknown Jan 10 at 5:17
• It is just divergence of harmonic series. You can search the net for 'divergence of harmonic series'. – Kavi Rama Murthy Jan 10 at 5:23
• A classic estimate: for each $n$, consider estimating $a_{2^n}$. – xbh Jan 10 at 5:26

• 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. – A. Arredondo Jan 10 at 9:16
$$a_1=1,a_4>2(=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{4}), a_8>\frac{5}{2}(=a_4+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}),a_{16}>3(=a_{8}+\frac{1}{16}+\frac{1}{16}+..8times)$$, 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.