Thank you for taking the time to reply. Please find attached the requested work. Please note the image starting 6E5 is intended to be read first.You stated that your conjecture was true for n=1. You need to show/prove that it is true for n=1. Why not just say that the theorem is true and be doee? Please show your work. Then I will look further on the work you posted
Thank you. I will take a look at this today. Much appreciate.You really need to know that 2N+1/2N = 2.
I think that you lost a 2 in N2<3N2.
You really need to know that 2N+1/2N = 2.
I think that you lost a 2 in N2<3N2.
You really need to know that 2N+1/2N = 2.
I think that you lost a 2 in N2<3N2.
Provo for anyone who tries to learn & use LaTeX. Thank you.I had a go writing it with latex.
\((a_n) = (\dfrac{n^2}{2^n})\)
Is this the correct way of stating that there is a sequence \(a_n\) and it consists of the terms \(\Big(\frac{n^2}{2^n}\Big)\) for each and every n where n is every natural number excluding 0?
I continue:
\(N\) is an integer such that if n \(\geq\) \(N\), then the ratio \(a_{n+1} / a_n < \frac{3}{4}\).
\(\dfrac{(n+1)^2}{2^{n+1}} * \dfrac{2^n}{n^2} < \dfrac{3}{4}\)
Thank you for the hint.Provo for anyone who tries to learn & use LaTeX. Thank you.
Here is a hint for fractions use dfrac{}{}, see above it gives larger displays.
Have a look HERE Is it clear that the limit is zero.
There is no need to clarify the question. Frankly I see little to no point assigning this in an algebra class.From looking at your table and by looking at a graph it appears that the sequence converges with zero as n tends to +infinity. Bingo
I think the questions is concerned with the nature of null sequences in some capacity because the following question asks, why is \(((\frac{3}{4})^na_N)\) a null sequence?There is no need to clarify the question. Frankly I see little to no point assigning this in an algebra class.
If you know calculus using l'Hopital's rule it is a easy proof:
\(\dfrac{{{n^2}}}{{{2^n}}}\mathop = \limits^H \dfrac{{2n}}{{{2^n}\log (2)}}\mathop = \limits^H \dfrac{2}{{{2^n}{{\left( {\log (2)} \right)}^2}}} \to 0\)
Why did you post the attachment sideways? It is not readable that way.I think the questions is concerned with the nature of null sequences in some capacity because the following question asks, why is \(((\frac{3}{4})^na_N)\) a null sequence? I have attached a picture of the question in full if you are interes
Sorry, here it is the right way up.Why did you post the attachment sideways? It is not readable that way.