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Does $\sum_{n=1}^{\infty} \frac{1}{n}$ converge? I actually do not know this, and have no steps to prove it, but my logic thinks that it is convergent. But when graphed, it takes ages for the function to "converge" to a single number. Thanks.

P.S. Sorry for the extremely amateur question

Xii
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    It in fact does not converge. – Sahiba Arora Jun 04 '17 at 13:14
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    Modulo the N vs x typo, it actually diverges, just very very slowly. It is called the harmonic series. – Ian Jun 04 '17 at 13:14
  • This is called the Harmonic Series, and if you're interested you can read about it on its corresponding wikipedia page. – Eff Jun 04 '17 at 13:27
  • Can anybody explain why this question is "bad"? – Xii Jun 04 '17 at 13:31
  • @Goodra: because the harmonic series is divergent, hence your question makes no sense at all. – Jack D'Aurizio Jun 04 '17 at 13:33
  • i don't think the downvotes are because it's amateur. It's because the question makes a statement that isn't obvious, (or true in this case) without explaining the reasoning behind it – jg mr chapb Jun 04 '17 at 13:37
  • @gebra could you elaborate? Maybe give an example – Xii Jun 04 '17 at 13:41
  • @JackD'Aurizio I'm an amateur / a 12 year old with practically no hope in mathematical reasoning ie Mathematics – Xii Jun 04 '17 at 13:43
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    It is definitely OK not knowing $\sum\frac{1}{n}$ diverges. However, the title of the question is problematic — if you don't know if it is convergent or not, how could you ask what number it converges to? "... my logic thinks that it is convergent to a number." This tells nothing. If you could provide more information about what is this magical "logic" that makes you think so, people would be more helpful to point out what is wrong with your reasoning. –  Jun 04 '17 at 13:56
  • The series is known to diverge, but quite interestingly the difference $\sum_{k=1}^n\frac1k-\log n$ converges to a constant known as $\gamma$ –  Jun 04 '17 at 14:10
  • Around the year $1300$ it was proved the $\mathbf{divergence}$ of the $\mathsf{Harmonic\ Series}$. – Felix Marin Jun 04 '17 at 18:04

2 Answers2

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This series diverges. The standard proof consists of swapping the terms for terms that are smaller, but easier to manipulate and which still makes the series diverge.

Take each term $\frac1n$, and swap the $n$ for the smallest power of $2$ that's not smaller than $n$. This makes $\frac12$ into $\frac12$, it makes $\frac13$ and $\frac14$ both into $\frac14$, it makes $\frac15, \frac16, \frac17$ and $\frac18$ all into four copies of $\frac18$, and so on. ($1$ stays as $1$, or is made into $\frac12$, you can choose.)

Now, note that each run of equal powers of $2$ sum up to $\frac12$. This means that the sum of all of them grows as large as we want it to. For instance, if we want to be certain that the sum is larger than $1000$, we could sum up $2000$ copies of $\frac12$, which means we need to sum all the way up to and including all the copies of $\frac{1}{2^{1999}}$.

And there is nothing special about $1000$ here. Pick any number, and I can pick a sum of these powers of $2$ that exceeds it. And in the original series, the terms are larger (or some times equal), so that must also be larger than the number you picked. Thus the series grows beyond any finite bound, and therefore diverges to infinity.

Arthur
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This series diverges, so it does not converge to a number.

Let's define:

$$H_n:=1+\frac 12+\cdots+\frac 1n.$$

Let's also define $u_n=H_n-\log n$ and $v_n=u_n-\frac 1n$.

You have $u_n-v_n>0$ and $u_n-v_n\to 0$.

You can prove easily that $(u_n)$ is decreasing since $u_n-u_{n+1}=-\frac1{n+1}-\log n+\log(n+1)=-\frac 1{n+1}-\log(1-\frac 1{n+1})\geqslant 0$ because $\log(1+x)\leqslant x$ for all $x$.

And similarly, $(v_n)$ is increasing.

So the sequences $(u_n)$ and $(v_n)$ are adjacent sequences, so they converge to a real number, let's call it $\gamma$ (it is called Euler's constant).

We have proven that:

$$\sum_{n=1}^N\frac 1n=\log N + \gamma +\text{ something that goes to $0$ as $N\to \infty$}.$$

E. Joseph
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