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Mathematics 25 Online
geerky42 (geerky42):

Determine if the series converge or diverge. \[\Huge\sum_{n=1}^{\infty} \dfrac{1}{2+\sqrt{n}}\]

geerky42 (geerky42):

I have a feel that I should use direct comparison, but I don't know how to...

OpenStudy (anonymous):

Well, to use direct comparison, you need a suspicion... what do you suspect of this series?

OpenStudy (anonymous):

I suspect it's divergent ^.^

OpenStudy (anonymous):

you can ignore 2 and i feel it would be converge

OpenStudy (anonymous):

if u imagine it as a continues function and finding the integral , you will find out tat it's infinity

OpenStudy (anonymous):

My favourite divergent series (and probably most anybody's) is \[\huge \sum_{n=1}^{\infty}\frac1n\]

geerky42 (geerky42):

So i should use integral test? @Roya

OpenStudy (anonymous):

so it seems that it's diverge

OpenStudy (anonymous):

No :/ It's easy to see (maybe using the Limit Comparison Test) that \[\huge \sum_{n=1}^{\infty}\frac1{2+n}\]

OpenStudy (anonymous):

i guess so, @PeterPan is right too

geerky42 (geerky42):

I tired this, for first four term, \(\dfrac{1}{n} \ge\dfrac{1}{2+\sqrt{n}}\) Should I worry about that?

OpenStudy (anonymous):

Don't use 1/n use its variant that I posted, 1/(2+n) It's easier to see it that way :)

OpenStudy (anonymous):

So, clearly \[\huge \frac{1}{2+n}\le\frac{1}{2+\sqrt n} \ \ \ \forall n \in \mathbb{N}\]

geerky42 (geerky42):

Ok thank!

OpenStudy (anonymous):

^.^

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