# Proof of the preliminary test (Divergence test for series)

1. Jun 19, 2013

### joej24

1. The problem statement, all variables and given/known data
$$Proove \, that \, if \, \lim_{n \to \infty} a_{n} \neq 0 \, , \, \sum^\infty a_{n} \, diverges$$

2. Relevant equations
$$S_{n} - S_{n-1} = a_{n}$$
$$\lim_{n \to \infty} S_{n} = S$$

3. The attempt at a solution
$$S_{n} - S_{n-1} = a_{n}$$
$$\lim_{n \to \infty} S_{n} - S_{n-1} = \lim_{n \to \infty} a_{n}$$
Since $a_{n} \neq 0$, we can divide by $a_{n}$

$$\lim_{n \to \infty} \frac {S_{n} - S_{n-1}} {a_{n}} = 1$$
We can rewrite $S_{n}$ and $S_{n-1}$ as $\sum^\infty a_{n}$ and $\sum^\infty a_{n-1}$

So, $$\lim_{n \to \infty} \frac {\sum^\infty a_{n} - \sum^\infty a_{n-1}} {a_{n}} = 1$$
Rearranging, $$\lim_{n \to \infty} \frac{\sum^\infty a_{n}} {a_{n}} - \sum^\infty \frac{a_{n-1}} {a_{n}} = 1$$
The term inside the second series is 1, so
$$\lim_{n \to \infty} \frac{\sum^\infty a_{n}} {a_{n}} = 1 + \sum^\infty 1$$

Thus, $$S_{n} = \lim_{n \to \infty} \sum^\infty a_{n} = a_{n} (1 + \sum^\infty 1)$$
Since $\sum^\infty 1$ diverges, we can say that $S_{n}$ diverges and therefore, if $\lim_{n \to \infty} a_{n} \neq 0$ , $\sum^\infty a_{n}$ diverges.

Is this proof complete?

Last edited: Jun 19, 2013
2. Jun 19, 2013

### LCKurtz

That isn't a correct statement of the theorem in the first place. When you say $\lim_{n \to \infty} a_{n} \neq 0$ you are implying $a_n$ has a limit which isn't $0$. That is not the correct denial of the statement that $a_n\rightarrow 0$. It may not converge to anything.

You can't take a limit since it isn't given $a_n$ has one. And nothing gives you $a_n\ne 0$.

I would suggest you try the showing the contrapositive: If $\sum a_n$ converges then $a_n\rightarrow 0$.