Proof of the preliminary test (Divergence test for series)

[joej24]

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?

 

[LCKurtz]

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 $a_n\rightarrow 0$.