El_Arrow
  • El_Arrow
test the series for convergence or divergence
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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El_Arrow
  • El_Arrow
\[\sum_{n=1}^{\infty} 1/(6n+1)\]
El_Arrow
  • El_Arrow
i used the comparison test but im not sure if its right
anonymous
  • anonymous
eyeball test the degree of the denominator is ?

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El_Arrow
  • El_Arrow
1
anonymous
  • anonymous
and the numerator?
El_Arrow
  • El_Arrow
it doesnt have one
anonymous
  • anonymous
of course it has one
El_Arrow
  • El_Arrow
is it 1
anonymous
  • anonymous
no
El_Arrow
  • El_Arrow
0
anonymous
  • anonymous
less
anonymous
  • anonymous
right
anonymous
  • anonymous
now we compare degrees \[1-0=1\]
anonymous
  • anonymous
since 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
  • El_Arrow
so we dont use any tests for this one
anonymous
  • anonymous
you 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
  • El_Arrow
but doesnt it converge when an is less than bn when using the limit comparison test
anonymous
  • anonymous
you 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
  • anonymous
yes, it does NOT converge again, you can see it with your eyeballs
anonymous
  • anonymous
degree 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
  • anonymous
oops i mean neither does \[\sum_{n=1}^{\infty}\frac{a}{bn+c}\]
El_Arrow
  • El_Arrow
so when i used the limit comparison test i got 1/6
anonymous
  • anonymous
you 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
  • anonymous
right and you compared it to a series that does NOT converge right?
El_Arrow
  • El_Arrow
yes 1/n
anonymous
  • anonymous
ok then since the limit is a finite non zero number, your series does not converge either
El_Arrow
  • El_Arrow
alright and the same thing goes for direct comparison test right
El_Arrow
  • El_Arrow
when its a finite non zero number it does not converge?
anonymous
  • anonymous
direct 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
  • El_Arrow
alright thanks
anonymous
  • anonymous
yw
El_Arrow
  • El_Arrow
oh one more question when does the integral test diverge or converge
El_Arrow
  • El_Arrow
@satellite73 @freckles
anonymous
  • anonymous
if the integral diverges, then so does the series and if the integral converges so does the series
El_Arrow
  • El_Arrow
but like does it diverge when its a finite non zero number or?
anonymous
  • anonymous
for 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
  • anonymous
the integral diverges, so does the sum
anonymous
  • anonymous
or perhaps i should say "the integral is infinite (does not exist) therefore neither does the sum"
anonymous
  • anonymous
to 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)
El_Arrow
  • El_Arrow
okay thanks
anonymous
  • anonymous
yw
anonymous
  • anonymous
you can still use a direct comparison: $$\sum_{n=1}^\infty \frac1{6n+1}=\sum_{n=2}^\infty\frac1{6(n-1)+1}=\sum_{n=2}^\infty\frac1{6n-5}$$now consider \(6n-5\le 6n\implies\frac1{6n-5}\ge\frac1{6n}\) and we know \(\sum_{n=2}^\infty\frac1{6n}=\frac16\sum_{n=1}^\infty\frac1n-\frac16\); clearly this diverges so it follows \(\sum_{n=2}^\infty\frac1{6n-5}\) diverges too
El_Arrow
  • El_Arrow
@oldrin.bataku what about for this one \[\sum_{n=1}^{\infty} \frac{ n^6+1 }{ n^7+1 }\]
El_Arrow
  • El_Arrow
can you use a comparison test like 1/n?
El_Arrow
  • El_Arrow
or the divergence test maybe?
anonymous
  • anonymous
$$\frac{n^6+1}{n^7+1}=\frac{1+1/n^6}{n+1/n^6}$$ 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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