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Mathematics 7 Online
OpenStudy (anonymous):

What is the difference between divergence and convergence?

OpenStudy (anonymous):

I can give you an example to explain if necessary

OpenStudy (amistre64):

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OpenStudy (amistre64):

divergence goes off into infinity; and convergence settles down to something that we can define

OpenStudy (amistre64):

determining "d" or "c" can be a pain tho

OpenStudy (anonymous):

wait just gotta show u something

OpenStudy (anonymous):

\[\int\limits_{1}^{\infty}1/x ^{2}\]

OpenStudy (amistre64):

convergent I believe; since it gets real small real quick

OpenStudy (anonymous):

and lets say \[\int\limits_{1}^{\infty}1/\sqrt{x}\]

OpenStudy (amistre64):

x^(-1/2) gets big

OpenStudy (amistre64):

i think i got that right lol

OpenStudy (anonymous):

ya u did lol

OpenStudy (anonymous):

but they are both getting smaller

OpenStudy (amistre64):

1/x is a dividing line; 1/x diverges, but anything just shy of it bigger, like x^(-1.00000000001) converges

OpenStudy (amistre64):

the sqrt doesnt do it fast enough

OpenStudy (amistre64):

1/x goes to gets smaller buth the sums never settle down

OpenStudy (anonymous):

oh ok

OpenStudy (amistre64):

the convergence is called a least upper boundary, not that that makes much difference. It just says that we can add up the partial sums and they tend towards a nice number or they dont

OpenStudy (anonymous):

oh ok get it now thanks :D

OpenStudy (amistre64):

youre welcome :)

OpenStudy (anonymous):

\[\int_a^{\infty} \frac{1}{x^r}dx=\int_a^{\infty}x^{-r}dx\] converges if \[r>1\] diverges if \[r\leq 1\]

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