Question

Calculus

Answer 1

\[\sum_{0}^{\infty} \frac{ 1 }{ (\sqrt 11)^2 }\]

Answer 2

n=0

Answer 3

Find the sum of the series, if it converges. Otherwise, enter DNE.

Answer 4

a bit confusing

Answer 5

most people would write
\[\frac{1}{11}\] rather than
\[\frac{1}{\sqrt{11}^2}\]

Answer 6

I'm sorry.., i mistyped.... that 2 is an "n"

Answer 7

and yes its confusing

Answer 8

\[\sum_{n=0}^{\infty} \frac{ 1 }{ (\sqrt 11)^n }\]

Answer 9

ooh ook

Answer 10

i did that when i wrote it down too lol! my mind does not like the n

Answer 11

use
\[\frac{1}{1-r}\] with
\[r=\frac{1}{\sqrt{11}}\]

Answer 12

isnt that after you know its convergent?

Answer 13

\[\frac{ 1 }{ 1-\sqrt 11 }\]

Answer 14

@Jlockwo3 This is actually a pretty fun formula to derive, do you want to spend a few minutes to learn it?

Answer 15

sure

Answer 16

it would help me a lot. yes please and thank you

Answer 17

i understand that it is \[\frac{ a }{ 1-r }\]

Answer 18

i'm sorry, i lost connection and did not realize it... did you get my replies?

Answer 19

Ok, so first off instead of thinking of each term as \[\Large \frac{1}{\sqrt{11}^n}\] we can put an n on the 1 as well since it won't affect it, then pull it off like this: \[\Large \left( \frac{1}{\sqrt{11}} \right)^n\] So that will be every term. Now you probably already know that if that value in there is less than 1 that it will converge. So maybe instead of 1/sqrt(11) we can replace this with the letter a just to make it easier to look at. \[\Large S=\sum_{n=0}^\infty a^n\] I'll call that total sum S. Ok so let's keep going...

Answer 20

ok... i understand the \[-1<r <1\]

Answer 21

If we write it out term for term we have: \[\Large S= a^0+a^1+a^2+a^3+a^4+ \cdots\] Now here's where it gets interesting. =) Multiply both sides of the equation by a and we get: \[\Large a*S=a^1+a^2+a^3+a^4+ \cdots\] Wait though, this looks very similar, if we just add a^0 (really anything to the zero power is 1) we will get this: \[\Large aS+a^0=a^0+a^1+a^2+a^3+a^4+ \cdots\] Now look at the right side of that equation. It's the original sum we had all along. Infinity lets us do weird stuff like this sometimes, so now we can just write "S" there on the right to get: \[\Large aS+a^0=S\] remember a^0=1 so we can replace it there, and solve for S we get: \[\Large aS+1=S \\ \Large 1=S-aS \\ \Large 1 = S(1-a) \\ \Large \frac{1}{1-a}=S\] feel free to replace a with 1/sqrt(11) :D

Answer 22

so, is it \[\frac{ 1 }{ 1-\sqrt 11 }\]

Answer 23

it will not be
\[\frac{1}{1-\sqrt{11}}\]though, it is
\[\frac{1}{1-\frac{1}{\sqrt{11}}}\]

Answer 24

oohhhh

Answer 25

this still looks yucky...

Answer 26

\[\frac{ 1 }{ \frac{ \sqrt 11 }{ \sqrt 11 }-\frac{ 1 }{ \sqrt 11 } }\]

Answer 27

multiply top and bottom by \(\sqrt{11}\) to clear the compound fraction

Answer 28

\[\frac{ 1 }{ 1-\frac{ \sqrt11 }{ 11 } }\]

Answer 29

\[\frac{ 1 }{ \frac{ 11 }{ 11 } -\frac{ \sqrt11 }{ 11 }}\]

Answer 30

\[\frac{ 1 }{ \frac{ 11 \sqrt11 }{ 11 } }= \frac{ 1 }{ \sqrt 11}\]

Answer 31

=\[\sqrt11\]

Answer 32

yea, im still not getting this lol