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El_Arrow
 one year ago
test the series for convergence or divergence
El_Arrow
 one year ago
test the series for convergence or divergence

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El_Arrow
 one year ago
Best ResponseYou've already chosen the best response.0\[\sum_{n=1}^{\infty} 1/(6n+1)\]

El_Arrow
 one year ago
Best ResponseYou've already chosen the best response.0i used the comparison test but im not sure if its right

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0eyeball test the degree of the denominator is ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0of course it has one

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now we compare degrees \[10=1\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0since the difference is one, the sum does NOT converge in order for a rational function to converge, the degree of the denominator has to be larger than the degree of the numerator by MORE than one anything more than one will do but one does not work

El_Arrow
 one year ago
Best ResponseYou've already chosen the best response.0so we dont use any tests for this one

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you can use a test if you like you could use the limit comparison test for example compare to \(\sum\frac{1}{n}\) which is known to diverge

El_Arrow
 one year ago
Best ResponseYou've already chosen the best response.0but doesnt it converge when an is less than bn when using the limit comparison test

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you cannot use the direct comparison test however since \[\frac{1}{6n+1}<\frac{1}{n}\] but the limit comparison test will work in general however what i said above is true

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes, it does NOT converge again, you can see it with your eyeballs

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0degree of the denominator has to be MORE than one larger then the degree of the numerator \[\sum\frac{1}{n}\] does not converge, neither does \[\sum\frac{a}{bx+c}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oops i mean neither does \[\sum_{n=1}^{\infty}\frac{a}{bn+c}\]

El_Arrow
 one year ago
Best ResponseYou've already chosen the best response.0so when i used the limit comparison test i got 1/6

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you can also use the integral test for this if you like you will get a log, and the log does not have a limit as \(n\to \infty\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0right and you compared it to a series that does NOT converge right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok then since the limit is a finite non zero number, your series does not converge either

El_Arrow
 one year ago
Best ResponseYou've already chosen the best response.0alright and the same thing goes for direct comparison test right

El_Arrow
 one year ago
Best ResponseYou've already chosen the best response.0when its a finite non zero number it does not converge?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0direct comparison you just compare if the terms are greater than a series that does not converge, NO if the terms are less than a series that does converge, YES but you can't really use it here

El_Arrow
 one year ago
Best ResponseYou've already chosen the best response.0oh one more question when does the integral test diverge or converge

El_Arrow
 one year ago
Best ResponseYou've already chosen the best response.0@satellite73 @freckles

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0if the integral diverges, then so does the series and if the integral converges so does the series

El_Arrow
 one year ago
Best ResponseYou've already chosen the best response.0but like does it diverge when its a finite non zero number or?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0for example you have \[\sum\frac{1}{6n+1}\] so \[a_n=\frac{1}{6n+1}\] put \[f(x)=\frac{1}{6x+1}\] and compute \[\int_1^{\infty}\frac{1}{6x+1}=\lim_{t\to \infty}\frac{1}{6}\log(6t+1)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the integral diverges, so does the sum

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0or perhaps i should say "the integral is infinite (does not exist) therefore neither does the sum"

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0to answer you question, NO if the integral is finite, then it exists (is a number) therefore the corresponding sum will be finite (not the same number as the integral, just finite)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you can still use a direct comparison: $$\sum_{n=1}^\infty \frac1{6n+1}=\sum_{n=2}^\infty\frac1{6(n1)+1}=\sum_{n=2}^\infty\frac1{6n5}$$now consider \(6n5\le 6n\implies\frac1{6n5}\ge\frac1{6n}\) and we know \(\sum_{n=2}^\infty\frac1{6n}=\frac16\sum_{n=1}^\infty\frac1n\frac16\); clearly this diverges so it follows \(\sum_{n=2}^\infty\frac1{6n5}\) diverges too

El_Arrow
 one year ago
Best ResponseYou've already chosen the best response.0@oldrin.bataku what about for this one \[\sum_{n=1}^{\infty} \frac{ n^6+1 }{ n^7+1 }\]

El_Arrow
 one year ago
Best ResponseYou've already chosen the best response.0can you use a comparison test like 1/n?

El_Arrow
 one year ago
Best ResponseYou've already chosen the best response.0or the divergence test maybe?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0$$\frac{n^6+1}{n^7+1}=\frac{1+1/n^6}{n+1/n^6}$$ consider that for \(n\ge1\) we have \(n+1\ge n+1/n^6\implies \frac1{n+1}\le\frac1{n+1/n^6}\le\frac{1+1/n^6}{n+1/n^6}\); since we know \(\frac1{n+1}\) diverges by an argument like the previous it follows that this series diverges too
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