Question

Hi everyone! Converge or Diverge via Root Test for Sigma n=1 to infinity of [(n+3)^2]/7^n...I posted the answer but need to discuss procedure...specifically, if you look at 26.bmp, what really bothers me is the part where it says
ln L = (1/7) times zero...I know you have to rid the left side of ln by introducing "e" on both sides, but what I don't u is why (1/&) seemingly dodges everything...My professor says this is correct but the procedure is very confusing...seems like I'm violating some rule of algebra...if anyone can discuss this with me, please do...Thanks! :o)

Answer 1

man that is a lot of work for not much bang for your buck

Answer 2

(1/&) should be (1/7)...sorry

Answer 3

your job is to compute
\[\lim_{n\to \infty}\sqrt[n]{a_n}\]

Answer 4

the denominator pretty clearly goes to \(7\) no agony there right?

Answer 5

any idea how you can introduce "e" and somehow not affect the 1/7 ?
It just feels really WRONG to me! :o(

Answer 6

Yeah it would seem to me you would get e^(0/7)=1. Seems funky to me.

Answer 7

forget the damn \(\frac{1}{7}\) it is just a number
worry only about
\[\lim_{n\to \infty}((n+3)^2)^{\frac{1}{n}}\]

Answer 8

Ah right, see you took out the 1/7 of the limit at the second step of the second row you drew.

Answer 9

I would multiply both sides by 7, then take the natural log of both sides

Answer 10

the question only is, what is
\[\lim_{n\to \infty}\left(n+3\right)^{\frac{2}{n}}\]

Answer 11

forget the \(\frac{1}{7}\) worry about it at the end

Answer 12

this is the old calculus problem of finding the indeterminate form \(\infty^0\)

Answer 13

okay satellite...i can ignore the 1/7, if you could just tell me procedurally, where I should store it...i can't just forget it...maybe you are way more advanced than myself, but I would really just like to know where you should technically put the 1/7 while everything else is going on?

Answer 14

so you would have

L = (1/7)*lim[ (n+3)^(2/n) ]

7L = lim[ (n+3)^(2/n) ]

ln( 7L ) = ln( lim[ (n+3)^(2/n) ] )

ln( 7L ) = lim[ ln( (n+3)^(2/n) ) ]

ln( 7L ) = lim[ (2/n) * ln(n+3) ]

Answer 15

just put it out of your mind for a moment and compute the above limit

Answer 16

_kind of_ look at what jim thompson is saying, he knows lol.

Answer 17

Yeah... i totally get what jim is saying...my professor though just waved his wand and magically never dealt with any of that, leaving us out in the cold to frigidly die a slow death by calculus!

Answer 18

what a cruel teacher lol...anyways, you found the limit to be 0, so this means

ln( 7L ) = lim[ (2/n) * ln(n+3) ]

ln( 7L ) = 0

7L = e^0

7L = 1

L = 1/7

Answer 19

of course you'll have your steps inserted in there, but you get the idea

Answer 20

if you want to compute that limit, you have a choice, you can rewrite it as
\[\left(n+3\right)^{\frac{2}{n}}=e^{\frac{2}{n}\ln(n+3)}\] and compute that limit, (which is what it really means anyway) or you can do that other thing
take the log, get
\[\frac{2}{n}\ln(n+3)\] then compute the limit, then exponentiate,which is a lot of chasing your tail, but what math teachers like to do

Answer 21

he kinda did what satellite is telling me to do...to just forget about it...but i can't forget about it...it is part of the problem and must be dealt with orderly and properly...geez

Answer 22

no offense to satellite or anything though! :o)

Answer 23

yes, but the limit of the denominator is 7 in any case!!

Answer 24

so your real job is to ascertain that
\[\lim(n+3)^{\frac{2}{n}}=1\] and you have a couple ways to do it

Answer 25

okay...i think I get it...thank you everyone! :o)

Answer 26

but keep going by all means...i enjoy learning different ways!

Answer 27

one is to take the log, then take the limit using either sheer obviousness or l'hopitals rule, then exponentiate (since your first step was to take the log)
the second and correct way is to write
\[(n+3)^{\frac{2}{n}}=e^{\frac{2}{n}\ln(n+3)}\] and then get
\[\lim_{n\to \infty}e^{\frac{2}{n}\ln(n+3)}=e^0=1\]

Answer 28

okay...I'm good! I got it!...question satellite...how did you get 100 on your smart score? I never say higher than 99 before lol

Answer 29

i kind of like "frigid death by calculus" though reminds me of "trickonomtry"
hope you don't mind if i use it

Answer 30

well...it's copyrighted, but I will give you a pass! :o)

Answer 31

i sent in the back of the match pack before anyone else, as well as two proof of purchase seals. that is all

Answer 32

"now get rid of the ln with e" that i don't like quite as much

Answer 33

, yeah, but you forgot to scratch off the grey stuff to reveal if you were a winner...i smell dispair in your future...lol

Answer 34

lol indeed

Answer 35

that'll cost you 30 points minimum! ha

Answer 36

ok...back to study...thanks again everyone!