Question

Someone help me with this sequence problem?

Answer 1

\[a _{n} = 3e ^{(n)/(2n+3)}\]

Answer 2

Show that this is monotonic/bounded and it's exact limit

Answer 3

I already can find the limit but its a matter of proving the monotonic/bounded part for all n that I've been having troubles with.

Answer 4

If we were to plug in n = 1, 2, 3, 4... we'd see that the exponent raises from 1/5 to 2/7 to 1/3 and so forth so that appears to be monotonic increasing and the limit is 3e^(1/2), so I know that the first term is the upper bound which is just 3e^(1/5). However, I need to prove this true for all n. Anyone have any suggestions?

Answer 5

@Kainui

Answer 6

I think consider \[a_n \overset{?}{<} a_{n+1}\] for n < some N

Answer 7

*n > some N... i.e. eventually monotonic

Answer 8

So you're suggesting that I use that an is less than or equal to an+1 and try to test n = 1,2,3 to see if that inequality holds true? So that i can say its monotonic increasing for n greater than or equal to 1?

Answer 9

true and show \[a_n>a_{n+1}\] for arbitrary n. don't use numbers

Answer 10

I had the sign the wrong way, i think it's monotonic and decreasing.

Answer 11

okay. so then that'd be:

\[3e ^{(n)/(2n+3)} \le 3e ^{(n+1)/(2n+5)}\]

Answer 12

and if it is. \[a_0\] is a bound

Answer 13

you want to show it... you don't know it yet... start with an equality

Answer 14

So then set that equal to eachother?

Answer 15

Sorry, I'm a bit lost.

Answer 16

um.. \[e^{(n+1)/(2n+5)}>e^{n/(2n+5)}...\] etc. I think it should work... havent done it in a while

Answer 17

oops, the sign is wrong again!!!! and i've gtg. sorry

Answer 18

thats alright. thanks for trying :p

Answer 19

@Directrix mind helping me out? :D

Answer 20

@imqwerty

Answer 21

What is the limit, as n approaches infinity, of n / (2n+3) ?

Answer 22

\[\begin{align}
b_{n+1} -{b_n}&=\dfrac{n+1}{2(n+1)+3}-\dfrac{n}{2n+3}\\~\\
&= \dfrac{(n+1)(2n+3)-n(2n+5)}{(2n+5)(2n+3)}\\~\\
&=\dfrac{3}{(2n+5)(2n+4)}\\~\\
&\gt 0
\end{align}\]

Answer 23

that shows \(b_n = \dfrac{n}{2n+3}\) is an increasing sequence

since \(e^x\) is an increasing function too, it follows that \(\large e^{b_n}\) is also increasing

Answer 24

ahhh alright. thanks you two was busy doing some other work. so we're actually subtracting the n+1th term of bn from the original bn?

Answer 25

\(b_{n+1}-b_{n}\gt 0 ~~\implies~~b_{n+1}\gt b_n\)

Answer 26

right. if we try plugging in values for n though for b(n+1) and b(n) we'd see that b(n+1) will always have the higher exponent so we'd use that inequality where bn+1 > bn and just subtract bn from both sides?

Answer 27

and by subtracting that bn from both sides that lets us know whether or not n is increasing or decreasing?

Answer 28

you can simply say

\[b_{n+1}\gt b_n~~\implies e^{b_{n+1}}\gt e^{b_n}\]

Answer 29

where \(b_n = \dfrac{n}{2n+3}\)

Answer 30

ok that makes sense. because they both have the same base you can just set it up for the exponent. thanks man. So that means n increases when greater than 0?

Answer 31

do you mean \(a_n\) is increasing ?

Answer 32

yes, my bad \[a _{n}\]

Answer 33

then it looks good

Answer 34

thanks!

Answer 35

yw