Ask your own question, for FREE!
Mathematics 8 Online
OpenStudy (anonymous):

first series is Summation from n=1 to inf 1/n and another series Summation from n=1to inf 1/n^2, why the first diverges and the second converges whereas both goes to zero

OpenStudy (anonymous):

not the both should converge?

OpenStudy (anonymous):

going to zero is "necessary but not sufficient"

OpenStudy (anonymous):

you can find an easy proof that \[\sum_{k=1}^n\frac{1}{k}\] can be made arbitrarily large by choosing large enough n

OpenStudy (anonymous):

i can give you the idea of the proof, but it is cumbersome to write it all here, easier simply to google it

OpenStudy (anonymous):

idea is take \[\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\] and say that it is less than \[\frac{1}{2}+\frac{1}{4}+\frac{1}{4}=1\] and keep going in this fashion, you can always add 1/2

OpenStudy (anonymous):

what should I google ? I mean if i just type the series as I did here then it doesn't lead me to any good website

OpenStudy (anonymous):

sorry, "greater than" not :less than

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

it is called the "harmonic series" and it is a well known divergent series

OpenStudy (anonymous):

so going to zero is certainly not enough, and this is the canonical example of a sequence that goes to zero, where the series does not converge

OpenStudy (anonymous):

whereas \[\sum\frac{1}{k^p}\] does converge if \[p>1\] but that is p must be strictly greater than 1

OpenStudy (anonymous):

integral test will prove this quickest

OpenStudy (anonymous):

understood

OpenStudy (anonymous):

what would I do without you hehehehe thanks bro

OpenStudy (anonymous):

are you mathematician?

OpenStudy (anonymous):

I mean you could be an engineer too?

OpenStudy (anonymous):

yw i am just satellite

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!