Question

Show that sigma of 1/n^2 IS CONVERGES by integral test

Answer 1

what are the limits with sigma ?

Answer 2

1 to infinity or 2 to infinity or other ?

Answer 3

using those limits, try to find
\(\int \dfrac{1}{x^2}dx\)
if the integral is finite, the sum converges,
if not the sum diverges

Answer 4

to infinity

Answer 5

from where to infinity ?
0 or 1 or 2 ?

Answer 6

from 1

Answer 7

ok, then try to solve
\(\huge \int \limits_{1}^{\infty} \dfrac{1}{x^2}dx\)

Answer 8

yes

Answer 9

can you solve ? do you know whats integral of x^-2 is ?

Answer 10

is it 2

Answer 11

nopes, how ?

Answer 12

I confused about that not sure

Answer 13

\(\int x^n dx=x^{n+1}/(n+1)+c\)
so what about integral of 1/x^2 = x^(-2)
??

Answer 14

i donot understand

Answer 15

have you ever solved an integral ?

Answer 16

no

Answer 17

then why are you using integral test ?

Answer 18

this new lesson for me

Answer 19

and i want to understand how it is work

Answer 20

i would suggest you to first try out easy examples of integration.
then come to this, you will understand only if you have solved some integration questions before.

Answer 21

i can slove it by other test but i want to understand integral test

Answer 22

for that you need to understand integration...

Answer 23

yes right

Answer 24

let me give you one example
integral of x^7 will be x^8/8 +c (in x^n, n=7, so n+1 =8)
using the formula i gave you above.
got this ?

Answer 25

oh yes i got it

Answer 26

so, what will be the integral of x^-2
hint: use n=-2

Answer 27

x^-1/-1

Answer 28

correct! you learn fast :)

so, integral of x^-2 is -1/x

now put the upper limit in it, that is put x = infinity in -1/x what do u get ?

Answer 29

do you mean x^-1/-1 i have to put x= infinity

Answer 30

or just for -1/x

Answer 31

yes.

the idea is . \(\int \limits_{lower \: limit }^{upper \: limit }f(x)dx = F(upper \:limit)-F(lower \: limit)\)

and x^-1/-1 and -1/x are SAME thing.

Answer 32

so put the upper limit in it, that is put x = infinity in -1/x what do u get ?

Answer 33

0
=0

Answer 34

good,
now put the lower limit in it, that is put x = 1 in -1/x what do u get ?

Answer 35

=-1

Answer 36

correct, so we have
\(\huge \int \limits_{1}^{\infty} \dfrac{1}{x^2}dx= -1/x|_{x=1}^{x=\infty}= 0-(-1)=1\)

got this ?
so we got the integral as FINITE.
what does we conclude from this ?

Answer 37

yes i got it now

Answer 38

integral FINITE means the sum converges :)

Answer 39

if its equal to one or lease than one mean converges

Answer 40

?

Answer 41

i have other question

Answer 42

thank you so much

Answer 43

sorry, i was away from keyboard, welcome ^_^
whats the other question ?

Answer 44

and for convergence by integral test,
we see whether the answer is FINITE or not, we don't compare it with 1
if finite, then converges