test the series for convergence or divergence

- El_Arrow

test the series for convergence or divergence

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- El_Arrow

\[\sum_{n=1}^{\infty} 1/(6n+1)\]

- El_Arrow

i used the comparison test but im not sure if its right

- anonymous

eyeball test
the degree of the denominator is ?

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## More answers

- El_Arrow

1

- anonymous

and the numerator?

- El_Arrow

it doesnt have one

- anonymous

of course it has one

- El_Arrow

is it 1

- anonymous

no

- El_Arrow

0

- anonymous

less

- anonymous

right

- anonymous

now we compare degrees
\[1-0=1\]

- 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

so we dont use any tests for this one

- 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

but doesnt it converge when an is less than bn when using the limit comparison test

- 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

yes, it does NOT converge
again, you can see it with your eyeballs

- 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

oops i mean neither does
\[\sum_{n=1}^{\infty}\frac{a}{bn+c}\]

- El_Arrow

so when i used the limit comparison test i got 1/6

- 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

right and you compared it to a series that does NOT converge right?

- El_Arrow

yes 1/n

- anonymous

ok
then since the limit is a finite non zero number, your series does not converge either

- El_Arrow

alright and the same thing goes for direct comparison test right

- El_Arrow

when its a finite non zero number it does not converge?

- 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

alright thanks

- anonymous

yw

- El_Arrow

oh one more question when does the integral test diverge or converge

- El_Arrow

@satellite73 @freckles

- anonymous

if the integral diverges, then so does the series and if the integral converges so does the series

- El_Arrow

but like does it diverge when its a finite non zero number or?

- 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

the integral diverges, so does the sum

- anonymous

or perhaps i should say "the integral is infinite (does not exist) therefore neither does the sum"

- 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

okay thanks

- anonymous

yw

- 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

@oldrin.bataku what about for this one \[\sum_{n=1}^{\infty} \frac{ n^6+1 }{ n^7+1 }\]

- El_Arrow

can you use a comparison test like 1/n?

- El_Arrow

or the divergence test maybe?

- 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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