without using the integral test directly show that the sum of(1/n) from n=1 to N diverges as N goes to infinity
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OpenStudy (anonymous):
hey 'whats up
OpenStudy (anonymous):
\[\sum_{n=1}^{N} (1/n) diverges as N \rightarrow \infty\]
OpenStudy (anonymous):
mathgirl
OpenStudy (anonymous):
whats up
OpenStudy (anonymous):
trying 2 figurure out this problem...:(
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OpenStudy (anonymous):
*figure
OpenStudy (anonymous):
AWW DONT CRY
OpenStudy (anonymous):
hey i know the answer
OpenStudy (anonymous):
what grade r u in mathgirl?
OpenStudy (anonymous):
really?
help me
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OpenStudy (anonymous):
k whats going on :P
OpenStudy (anonymous):
in in uni, 2nd year
OpenStudy (anonymous):
i am to
OpenStudy (anonymous):
maybe we can hang out and have sex
OpenStudy (anonymous):
thats so imature dud..
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OpenStudy (anonymous):
freak!!
OpenStudy (anonymous):
lol shut up andrew garcia
OpenStudy (anonymous):
get lost!
OpenStudy (anonymous):
If you don't need anything major as a proof, you could simply state that p series only converge if s > 2, and seeing as s = 1 in this problem, s < 2, therefore it does not converge. If you want a more complicated proof:
http://www.math.unh.edu/~jjp/proof/proof_n.html
OpenStudy (anonymous):
wow that guy is a freak ..
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OpenStudy (anonymous):
i was jk !!!!!!!!!!!!!!!
OpenStudy (anonymous):
hey mathsgirl whats the problem
OpenStudy (anonymous):
its at the top...i think i should use the comparison test but im not sure how
OpenStudy (anonymous):
mathsygirl im sorry !
OpenStudy (anonymous):
really
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OpenStudy (anonymous):
il help you get the answer mathsgirl
OpenStudy (anonymous):
thanks qwer the link is quite good :)
OpenStudy (anonymous):
jonathan2...really, get lost!
OpenStudy (anonymous):
woow!!!!!!!!!!!!
OpenStudy (anonymous):
your nice
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OpenStudy (anonymous):
You can use the comparison test by testing another divergent series. If that divergent series is always less than 1/n then that implies 1/n is divergent as well.