Question

test the series for convergence or divergence

Answer 1

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

Answer 2

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

Answer 3

eyeball test
the degree of the denominator is ?

Answer 4

1

Answer 5

and the numerator?

Answer 6

it doesnt have one

Answer 7

of course it has one

Answer 8

is it 1

Answer 9

no

Answer 10

0

Answer 11

less

Answer 12

right

Answer 13

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

Answer 14

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

Answer 15

so we dont use any tests for this one

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

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
\[\sum\frac{a}{bx+c}\]

Answer 21

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

Answer 22

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

Answer 23

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 \(n\to \infty\)

Answer 24

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

Answer 25

yes 1/n

Answer 26

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

Answer 27

alright and the same thing goes for direct comparison test right

Answer 28

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

Answer 29

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

Answer 30

alright thanks

Answer 31

yw

Answer 32

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

Answer 33

@satellite73 @freckles

Answer 34

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

Answer 35

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

Answer 36

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
\[\int_1^{\infty}\frac{1}{6x+1}=\lim_{t\to \infty}\frac{1}{6}\log(6t+1)\]

Answer 37

the integral diverges, so does the sum

Answer 38

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

Answer 39

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)

Answer 40

okay thanks

Answer 41

yw

Answer 42

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

Answer 43

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

Answer 44

can you use a comparison test like 1/n?

Answer 45

or the divergence test maybe?

Answer 46

$$\frac{n^6+1}{n^7+1}=\frac{1+1/n^6}{n+1/n^6}$$
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