ratio test for series

[karmic orbit]

the series converges but the answer i got for the limit is wrong. anyone know where i went wrong?

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[frail sable]

huh?

in the ratio test if youre getting <1 it converges and you are getting <1

[karmic orbit]

i know

ik it doesnt matter much but the limit equals 1/2 somehow

[frail sable]

0<1 so it converges then

[karmic orbit]

so im jw where i went wrong

[frail sable]

ah ah

hm your steps arent clear

[karmic orbit]

which step

[frail sable]

after the second step u can for simplicity just take the denominators reciprocal and multiply it

[karmic orbit]

thats what the 3rd step is

then i got a bunch of cancelations

then foiled the remaining

then i used l'hopital's

[frail sable]

so the problem is in between the second and fourth step

upon simplifying i am getting $\frac{n+3}{2(n+1)}$ which does converge to 1/2 as $n$ tends to infinity, which is also easily seen without L'Hopital (however with L'Hopital it is seen to be 1/2 too)

[sterile light]

wolfqz

[frail sable]

let me show u the simplification

[karmic orbit]

how does the (n+2) cancel

[frail sable]

$\frac{(n+3)!}{2!(n+1)!2^{n+1}} \times \frac{2!n!2^n}{(n+2)!}$

$\frac{(n+3)(n+2)!}{(n+1)!2^n} \times \frac{2^nn!}{(n+2)!}$

For the second step, i just wrote (n+3)! using the recursive formula
cancelled out 2!
wrote 2^(n+1) as 2^n times 2

[karmic orbit]

ohhh i simplified (n+3)! into (n+3)n!

[sterile light]

wolfqz

[frail sable]

right, so are you getting the correct answer now?

[karmic orbit]

yes