Does $a_n = \frac{1}{n}$ converges even though $\sum \frac{1}{n}$ diverges?

Question

I've been studying sequences and series recently. As I understood, the sequence convergence is determined whether the sequence has a limit value. Now, in this example

$$a_n = \frac{3n^2 - 5n + 7}{3n^3 - 5n + 7}$$

I get $\lim_{n\to \infty} \frac{1}{n} = 0$, which means this sequence converges. What confuses me is that it is known that series $\sum \frac{1}{n}$ diverges. My question is, is it possible that $\frac{1}{n}$ converges when working with sequences, but diverges when working with series? Or it diverges in both cases?

Thank you in advance

Answer 1

If a series converges, its general term must converge to zero. Indeed, if $s_{0} = 0$, we have: \begin{align*} s = \sum_{n=1}^{\infty}a_{n} \Longrightarrow s_{n} - s_{n-1} = a_{n} \Longrightarrow \lim_{n\rightarrow\infty} a_{n} & = \lim_{n\rightarrow\infty} (s_{n}-s_{n-1})\\ & = \lim_{n\rightarrow\infty} s_{n} - \lim_{n\rightarrow\infty} s_{n-1} = s - s = 0 \end{align*}

However, the converse is not true. As you have noticed, according to the integral criterion, the harmonic series diverges \begin{align*} \int_{1}^{\infty}\frac{\mathrm{d}x}{x} = +\infty \Longrightarrow \sum_{n=1}^{\infty}\frac{1}{n} = +\infty \quad \text{but}\quad\lim_{n\rightarrow\infty}\frac{1}{n} = 0 \end{align*}

Is this what you are asking for?

Answer 2

Suppose you start walking, and the first step is one meter long. The next two steps are half a meter long. The three steps after that are each one third of a meter long. And so on. The length of your steps get smaller and smaller, and converge to zero. But the total distance you've traveled doesn't converge; it just keeps increasing without bound. It's possible for your speed to converge to zero, without the total distance traveled converging.

Answer 3

$\frac 1n$ is $\frac 1n$.

$\sum_{k=1}^n \frac 1k$ is $1 + \frac 12 + \frac 13 + ....... + \frac 1n$.

Those are completely different things.

My question is, is it possible that $\frac 1n$ converges when working with sequences, but diverges when working with series?

That question doesn't really make any sense. "$\frac 1n$" doesn't do anything and isn't a thing that converges. A sequence (a set listed in order) of $\{1, \frac 12, \frac 13,....\} = \{a_k| a_k = \frac 1k \forall k \in \mathbb N\}$ may converge. And $\sum_{k=0}^{\infty} a_k = \sum_{k=0}^{\infty} \frac 1k = 1 + \frac 12 + \frac 13 + .....$ may diverge. This is because $\{1, \frac 12, \frac 13,....\} = \{a_k| a_k = \frac 1k \forall k \in \mathbb N\}$ and $\sum_{k=0}^{\infty} a_k = \sum_{k=0}^{\infty} \frac 1k = 1 + \frac 12 + \frac 13 + .....$ are two entirely different things.

This is akin to $57 + 39 + 6$ is a different thing than $57,39, 6$.

It's pretty easy to convince oneself that $\lim_{n\to \infty} \frac 1n = 0$. The terms $\frac 1n$ get increassingly smaller but no matter how small a value we compare the terms to we can find a $\frac 1n$ term that is smaller.

Or in other more technical terms. For any $\epsilon > 0$ we can find an $n > \frac 1\epsilon$ where $0 < \frac 1n < \epsilon$ and that is true for any $\epsilon$ no matter how small but positive.

It's less obvious that $\sum_{k=1}^{\infty} \frac 1k$ diverges but this is a sum and we keep adding on to it. It doesn't matter that the terms we keep adding on might converge down to something as we are considering the sum of terms; not just the current term we are adding.

In this case although the individual terms "go to zero", they don't go to zero "fast enough". $\sum_{k=1}^{\infty} \frac 1k = 1 + (\frac 12 + \frac 13) + (\frac 14 + \frac 15 + \frac 16 + \frac 17) + ...... + (\frac 1{2^n} + \frac 1{2^n + 1} + .... + \frac 1{2^n + (2^n -1)}) + ..... > \frac 12 + (\frac 14 + \frac 14) + (\frac 18 + \frac 18 + \frac 18 + \frac 18) + ...... (\frac 1{2^{n+1}} + \frac 1{2^{n+1}} + .... + \frac 1{2^{n+1}}) + ... = \frac 12 + \frac 12 + .... + \frac 12 + .....$ which clearly diverges.