Question

Ratio Test problem:
Summation {n=1}^infty (n/(2n+5))^n
I got L = 1/(2e^2) but I don't know where/how else to check my answer. I see that it converges since 1/(2e^2) <1.. but that's it. Any help is appreciated! Thank you :)

Answer 1

yeah

Answer 2

well @dan815 lets get this young lady a A+++++++++++++++++

Answer 3

\[\sum_{1}^{\infty} (\frac{ n }{ 2n+5 })^{n}\]
\[L = \frac{ 1 }{ 2e^2 }\]

Answer 4

are you given 1/2e^2 or did u find out that it is less than that bound

Answer 5

I'll assume the question is
10n2−1+2n−5n−1=2n+5n+1
eliminate the denominator by multiplying every term by (n-1)(n+1)
10+(2n−5)(n+1)=(2n+5)(n−1)
simplify the expression
10+2n2−3n−5=2n2+3n−5
collect like terms gives
10−6n=0
so 6n = 10 n = 5/3

Answer 6

I got 1/2e^2

and, give me a min to read that last post, haha

Answer 7

ratio test this converges

Answer 8

wait wat wat i did did not help u at all X_X

Answer 9

so i went on wolframalpha to double check my answer, and their decimal approximation was larger than mine.. so yeah i don't know where i went wrong.. the three terms (after taking the limit) came out to be e*e^3*(1/2)

Answer 10

oh you want to evaluate this?

Answer 11

yes @dan815 and you did @TheRealMeeeee i just went about solving for it slightly differently sorry!

Answer 12

oh

Answer 13

probably did it wrong.. it got kind of messy with everything raised to the n.. i wanted to perform the root test and then the ratio test, but i'm pretty sure that's not legal in math

Answer 14

well dan should get the medal dont give me one give him one

Answer 15

show me your work

Answer 16

brb dan

Answer 17

ok, i started with
1. |((n+1)/(2n+7))^(n+1)| divided by a_n

2. I combined like terms (as best as i could):
| ((n+1)^n)(n+1)(2n+1)^n | divided by | ((2n+7)^n)((2n+7)(n^n) |

3. simplified so i could get the limit at n-> inf
| (((n+1)/n)^n)(((2n+1)/(2n+7))^(n+1))((n+1)/(2n+7))

Answer 18

wait, that's the comparison test though..?

Answer 19

and that converges to...0

Answer 20

Well it doesnt converge to 0 but ya it does converge

Answer 21

what would it converge to then? i thought the lim(n->inf) (1/2n)^n is the same as (1/inf)^(inf) = 0^inf = 0

Answer 22

the sum wont converge

Answer 23

also, does the work i showed you make sense or is it unclear? writing in equation is too long

Answer 24

bk bro

Answer 25

yep looks fine

Answer 26

so rember always ask the smart crew me and @dan815

Answer 27

i didn't even know i could ask specific people questions.. i just post these and let any random person ask haha.. i will definitely keep you guys posted in the future. im sure youlove calc 2

Answer 28

lol good