Question

help? @jim_thompson5910

Answer 1

with what? :)

Answer 2

Hold your horses, I'm posting it. :)

How do you use the comparison test? I think it is convergent though.

Answer 3

sorry wrong file

Answer 4

lol okay i was like..what the...?

Answer 5

what do u think? and how u get the answer that u think?

Answer 6

cause the common ratio is less than 1, so convergent

Answer 7

|r| < 1, convergent.

but i don't know how to do the comparison test

Answer 8

it is 1/m^2
1/7^2 + 1/8^2 and so on

Answer 9

1/m^2 has a limit of zero...does that have anything to do with it?

Answer 10

We can rewrite our series as 1/7^2 + 1/8^2 + 1/9^2 which is the same as sum from 7 to infinity of 1/n^2

Answer 11

okay...is that the comparison test...?

Answer 12

which we can compare to the harmonic series 1/n, which is divergent
what can we say about the terms 1/n^2 and 1/n, when n is large?

Answer 13

\(\color{blue}{\text{Originally Posted by}}\) @StudyGurl14
okay...is that the comparison test...?
\(\color{blue}{\text{End of Quote}}\)
That's what I was asking, do you remember what teh test says?

Answer 14

See this link

http://tutorial.math.lamar.edu/Classes/CalcII/IntegralTest.aspx

specifically, see what I've attached (which is found on that page)

Answer 15

sorry, openstudy rebooted annoyingly while i was typing something

Answer 16

Same here :'(

Answer 17

\(\huge\color{Orange}{\rightarrow\succ}\cal\color{Blue}{\bigstar\color{orange}{ JOHNKNOWS}\bigstar}\color{Orange}{\prec\leftarrow}\)

Answer 18

so, since p=2, it converges, right @jim_thompson5910 ?

Answer 19

No, that is not the reason. First, see if we can use the comparison test. Are the conditions fulfilled?

The test says that if we have two series with the terms positive
and if the terms of the first series are all smaller than or equal to the terms of the second series, then if the second series is convergent, so will the first series be
if the first series is divergent, so is the second series

Answer 20

(1) We are comparing our series 1/n^2 to 1/n all of the terms are positive in both.
(2) 1/n^2 is always smaller than 1/n. (for n greater than 7 of course)

So both conditions are fulfilled, which means we can use the comparison test:
because 1/n, the harmonic series, diverges, 1/n is divergent but 1/n^2 is convergent

Think like this:
if the "larger" series converges, so does the smaller one; if the smaller series diverges, so does the larger one. If the larger one diverges, the smaller one converges

Answer 21

@StudyGurl14 yes correct. Since p = 2 makes p > 1 true, the infinite series 1/n^2 converges due to the p-series test.

I'm not sure which sequence we can look at where each term is larger than 1/n^2 to use the comparison test, but the comparison test isn't really needed. You can just use the p-series test.