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## Homework Statement

[tex]Proove \, that \, if \, \lim_{n \to \infty} a_{n} \neq 0 \, , \, \sum^\infty a_{n} \, diverges[/tex]

## Homework Equations

[tex] S_{n} - S_{n-1} = a_{n}[/tex]

[tex] \lim_{n \to \infty} S_{n} = S[/tex]

## The Attempt at a Solution

[tex] S_{n} - S_{n-1} = a_{n}[/tex]

[tex] \lim_{n \to \infty} S_{n} - S_{n-1} = \lim_{n \to \infty} a_{n}[/tex]

Since [itex] a_{n} \neq 0 [/itex], we can divide by [itex]a_{n}[/itex]

[tex] \lim_{n \to \infty} \frac {S_{n} - S_{n-1}} {a_{n}} = 1[/tex]

We can rewrite [itex] S_{n} [/itex] and [itex] S_{n-1} [/itex] as [itex] \sum^\infty a_{n} [/itex] and [itex] \sum^\infty a_{n-1} [/itex]

So, [tex] \lim_{n \to \infty} \frac {\sum^\infty a_{n} - \sum^\infty a_{n-1}} {a_{n}} = 1[/tex]

Rearranging, [tex] \lim_{n \to \infty} \frac{\sum^\infty a_{n}} {a_{n}} - \sum^\infty \frac{a_{n-1}} {a_{n}} = 1[/tex]

The term inside the second series is 1, so

[tex] \lim_{n \to \infty} \frac{\sum^\infty a_{n}} {a_{n}} = 1 + \sum^\infty 1 [/tex]

Thus, [tex] S_{n} = \lim_{n \to \infty} \sum^\infty a_{n} = a_{n} (1 + \sum^\infty 1) [/tex]

Since [itex] \sum^\infty 1[/itex] diverges, we can say that [itex] S_{n} [/itex] diverges and therefore, if [itex]\lim_{n \to \infty} a_{n} \neq 0 [/itex] , [itex] \sum^\infty a_{n} [/itex] diverges.

Is this proof complete?

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