Proof of the preliminary test (Divergence test for series)

Homework Statement

$$Proove \, that \, if \, \lim_{n \to \infty} a_{n} \neq 0 \, , \, \sum^\infty a_{n} \, diverges$$

Homework Equations

$$S_{n} - S_{n-1} = a_{n}$$
$$\lim_{n \to \infty} S_{n} = S$$

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:

Related Calculus and Beyond Homework Help News on Phys.org
LCKurtz
Homework Helper
Gold Member

Homework Statement

$$Proove \, that \, if \, \lim_{n \to \infty} a_{n} \neq 0 \, , \, \sum^\infty a_{n} \, diverges$$
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.

Homework Equations

$$S_{n} - S_{n-1} = a_{n}$$
$$\lim_{n \to \infty} S_{n} = S$$

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 $a_{n}$
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##.