determine whether n/[2^(n+2)] is bounded

Question

chrisplusian · Answer

@amistre64 I don't know why these are giving me such a hard time. If you can spare a moment I would appreciate you following my progression to see where I am going wrong.

chrisplusian · Answer

$$A_{n}=\frac{ n }{ 2^{n+2} }$$

chrisplusian · Answer

$$0<\frac{ 1 }{ 2^{n+2} } <\frac{ 1 }{ 2^{n+1} }$$

chrisplusian · Answer

$$0<\frac{ n }{ 2^{n+2} } <\frac{ n }{ 2^{n+1} }$$

chrisplusian · Answer

$$0<n(2^{n+2})<n(2^{n+1})$$

chrisplusian · Answer

and from there I am lost

amistre64 · Answer

im no expert, but why did you choose <1/2^(n+1)  instead of <1/4  or <1/2 ?

chrisplusian · Answer

Well because this deals with sequences so I know n has to be an integer. I Have to find the least upper and least lower bound, meaning the integer closest to what it is bound between. Because two is in the denominator I know for sure that 2^n+1 is a smaller number than 2^n+2. I am not saying that it is the best reasoning, but like I said this concept is giving me a hard time.

chrisplusian · Answer

This makes the 1/(2^{n+1}) a larger number than 1/(2^{n+2})

chrisplusian · Answer

And as I said this is just my thought process so I am trying to figure out what I need to correct

amistre64 · Answer

my idea is ore to do with getting rid of the "n" in the exponent of the comparison altegether
$$\frac{n}{2^{n+2}}=\frac{n}{4~2^n}<\frac n{4}$$

chrisplusian · Answer

Ok where do you go from there because I tried that too and then I multiplied the inequality through by the lcd and it gave me:
$$0<n<n(2^n)$$

amistre64 · Answer

wish i knew :)  im gonna have to read up on this to refresh the memory

chrisplusian · Answer

Ok thanks

amistre64 · Answer

good luck, my computers being way too slow for this place today :/

chrisplusian · Answer

@Mertsj do you know how to do this?

chrisplusian · Answer

@Hero ?

chrisplusian · Answer

@.Sam. ?

mertsj · Answer

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