anonymous
  • anonymous
What is the difference between divergence and convergence?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
I can give you an example to explain if necessary
amistre64
  • amistre64
|dw:1327426598793:dw|
amistre64
  • amistre64
divergence goes off into infinity; and convergence settles down to something that we can define

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

amistre64
  • amistre64
determining "d" or "c" can be a pain tho
anonymous
  • anonymous
wait just gotta show u something
anonymous
  • anonymous
\[\int\limits_{1}^{\infty}1/x ^{2}\]
amistre64
  • amistre64
convergent I believe; since it gets real small real quick
anonymous
  • anonymous
and lets say \[\int\limits_{1}^{\infty}1/\sqrt{x}\]
amistre64
  • amistre64
x^(-1/2) gets big
amistre64
  • amistre64
i think i got that right lol
anonymous
  • anonymous
ya u did lol
anonymous
  • anonymous
but they are both getting smaller
amistre64
  • amistre64
1/x is a dividing line; 1/x diverges, but anything just shy of it bigger, like x^(-1.00000000001) converges
amistre64
  • amistre64
the sqrt doesnt do it fast enough
amistre64
  • amistre64
1/x goes to gets smaller buth the sums never settle down
anonymous
  • anonymous
oh ok
amistre64
  • 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
anonymous
  • anonymous
oh ok get it now thanks :D
amistre64
  • amistre64
youre welcome :)
anonymous
  • anonymous
\[\int_a^{\infty} \frac{1}{x^r}dx=\int_a^{\infty}x^{-r}dx\] converges if \[r>1\] diverges if \[r\leq 1\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.