Question

test the series for convergence or divergence. Use comparison or limit comparison test.

Answer 1

\[\sum_{n=1}^{\infty} \frac{n }{ n(n+2) }\]

Answer 2

do i do partial fraction decomposition here?

Answer 3

whats stopping you from doing the obvious.. cancelling n top and bottom ?

Answer 4

nothing

Answer 5

just double check there are no typoes

Answer 6

\[\sum_{n=1}^{\infty} \frac{ 1 }{ n+2 }\]

Answer 7

\[an = \frac{ 1 }{ n+2 }.....bn \frac{ 1 }{ n}\]

Answer 8

i mean bn = 1/n

Answer 9

looks good, keep going..

Answer 10

\[\frac{ 1 }{ n+2 } > \frac{ 1 }{ n }......diverges?\]

Answer 11

that inequality doesn't look correct

n+2 > n

that means

1/(n+2) < 1/n

Answer 12

yeah and if we plug in for example n=3......(1/5)<(1/3) so it converges

Answer 13

right?

Answer 14

nope, SUM 1/n doesn't converge.

Answer 15

it doesnt? why not?

Answer 16

\(\sum\limits_{n=1}^{\infty} \dfrac{1}{n}\) is harmonic series

Answer 17

oh yes it is harmonic

Answer 18

so then we can use either directive or limit comparison test right?

Answer 19

yes use limit comparison test may be

Answer 20

\(a_n = \dfrac{1}{n+1}\) and \(b_n = \dfrac{1}{n}\)

\[\lim\limits_{n\to\infty}\dfrac{a_n}{b_n} = ?\]

Answer 21

sorry i lost connection

Answer 22

\[\lim_{n \rightarrow \infty} \frac{ \frac{ 1 }{ n+2 } }{ \frac{ 1 }{ n } } = \frac{ n }{ n+2 }\]

Answer 23

yes what does that evaluate to

Answer 24

\[\frac{ \infty }{ \infty }\]

Answer 25

use lhopital rule if u have to

Answer 26

then it will be \[\lim_{n \rightarrow \infty} \frac{ \ln n }{ n+2 } \]

Answer 27

from where are you getting logarithms ?

Answer 28

Div

Answer 29

\[\frac{ \frac{ n }{ n } }{ \frac{ n }{ n }}+\frac{ \frac{ n }{ n } }{ \frac{ 2 }{ n } }\]

Answer 30

is that better?

Answer 31

\[\lim_{n \rightarrow \infty} \frac{ \frac{ 1 }{ n+2 } }{ \frac{ 1 }{ n } } = \lim_{n \rightarrow \infty} \frac{ n }{ n+2 } = \lim_{n \rightarrow \infty} \dfrac{1}{1} = 1\]

Answer 32

so l'hopital's rule is just like taking the derivative?

Answer 33

yes
http://mathworld.wolfram.com/LHospitalsRule.html

Answer 34

i kindy forgot how to do it hehehe....

Answer 35

i got this other problem its the last one \[\sum_{n=1}^{\infty} \frac{ 1+\sqrt{n} }{ n}\]

Answer 36

just want to see if i do it right

Answer 37

\[an = \frac{ 1+\sqrt{n} }{ n }.....bn=\frac{ \sqrt{n} }{ n }=\frac{ n^\frac{ 1 }{ 2 } }{ n }=n^\frac{ 1 }{ 2 }\]

Answer 38

use direct comparison test with harmonic series maybe..

Answer 39

for \(n\gt 0 \) we have \[\dfrac{1+\sqrt{n}}{n} \gt \dfrac{1}{n}\]

since the harmonic series \(\sum \dfrac{1}{n}\) diverges, the series \(\sum \dfrac{1+\sqrt{n}}{n}\) also diverges/

Answer 40

how do you know to use direct comparison test with harmonic series?

Answer 41

do you see that

\(\frac{1+\sqrt{n}}{n}\) is greater than \(\frac{1}{n}\) for all \(n\gt 0\) ?

Answer 42

oh yes so thats how you test it to see if its harmonic

Answer 43

good, but we're not testing if it is harmonic or not

we're just showing the terms in our series are GREATER than the terms in harmonic seris

Answer 44

so if its greater we use direct comparison test