# Why does the sequence $\frac{1}{n}$ diverge?

My math teacher said that it is not a Cauchy sequence thus it isn't convergent but divergent. It is bounded and its limit exists ($=0$). Now i am really confused. I thought that things started to get clearer and I then came across this one.

She also said that $\frac{1}{n^2}$ is convergent. I am not confusing the terms "sequence" with "series". Thanks a lot.

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The series $\sum \frac1n$ diverges. The sequence $\frac1n$ converges. –  Daniel Fischer Oct 11 at 17:29
If you really aren't confusing between sequence and series and if your teacher really said what you said he did then he commited a big blunder. –  DonAntonio Oct 11 at 17:31
there is more, i am in college... –  Tamás Oct 11 at 17:32
if $a_{n}=\sum_{k=1}^{n}\frac{1}{k}$ then $a_{n}$ diverges. –  drhab Oct 11 at 17:32

If the question is "What did my teacher mean to say?" (putting aside the issue of what they actually did say) I'd guess:

The sequence (of partial sums) $$s_{n} = \sum_{k=1}^n \frac{1}{k}$$ is divergent (in fact, unbounded) and hence not Cauchy, but $$|s_{n+1} - s_{n}| = \frac{1}{n +1} \to 0$$ (i.e., the sequence of summands is bounded, and the limit of the summands is zero). It's entirely possible your instructor (correctly) said "sequence" in reference to both examples.

The sequence of summands $(a_n) = (\frac{1}{n})$ is (as you, Daniel Fischer, and others point out) obviously convergent to $0$, and obviously Cauchy from the definition (since $|a_n - a_{n+m}| < \frac{1}{n}$ for all $n$ and $m > 0$). The post by abiessu shows the sequence $(s_n)$ of partial sums is not Cauchy.

In my experience, the pronoun "it" causes confusion when discussing convergence of sequences and series. In fact, "it" is a paved walkway to mathematical miscommunication, and is best avoided when feasible. (If you can't replace "it" with a noun, there's something you don't fully understand. :)

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One reason that the series diverges is as follows. Group terms together in the following way:

\begin{align} a_n&=\sum_{k=1}^{2^n}\frac 1k\\ &=(1)+(1/2)+(1/3+1/4)+(1/5+1/6+1/7+1/8)+\cdots+(1/(2^{n-1}+1)+1/(2^{n-1}+2)+\cdots+1/(2^n-1)+1/2^n)\\ &\ge\frac 1 2 + \frac 1 2 + \frac 1 2 + \frac 1 2 + \cdots +\frac 1 2 \end{align}

This series is then a sum of $n+1$ terms each of which is greater than or equal to $\frac 12$, and we can always construct another term. Then we have:

$$\lim_{n\to\infty}a_n=\infty$$

The above reasoning is how we know that the series diverges. Why it diverges is a very deep question. In general, for all $r\gt 1$, the series $\sum_{k=1}^\infty\frac 1{k^r}$ converges. In an unusual twist, $\sum_p\frac 1p$ the sum over the reciprocal of the primes also diverges.

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You really should change the "up to infinity" on the summation to a finite value. It's incorrect (paradoxical) to arbitrarily group infinite sums. Arbitrarily grouping finite sums is fine though. You might just be trying to keep the argument simple, but since it's someone new to sums, it's best if he doesn't get a misunderstanding about grouping. –  DanielV Oct 11 at 18:01
@DanielV: good point, how does the updated version look? –  abiessu Oct 11 at 18:38
This does the grouping correctly and still makes the point that for any upper limit you chose, there is always an $a_n$ that exceeds it. –  DanielV Oct 11 at 20:49