Properties of Limits-At infinity

[ku1005]

Hey all, just wndering with the following question, what would be the best method to appraoch the question?

"Q) Using properties of limits, show that there exists an N in the set of Positive Integers such that :

"

My attempt was to firstly show that the sequence is decreasing, ie



Thenm solve for when {((n+1)^2)/(2n^2)} - 3/4 = 0

And let N = this value of n solved.

However, Im sure there is a much easier way to do these, but I have forgotten, I seem to remember expressing the sequence as something simpler, solving then for when this simpelr sequence = 3/4, as hence the more complex sequence is also less at this value.

ANY tips/hints would be greatly appreciated!

cheers

 

[ku1005]

Actually, my limit calculation is wrong via that method,

howeve checking using an alternative method

ie if decreasing then satisfies

lim as n_> infinity for

a(n+1) - a(n) <1

the sequence is definitely decreasing, sorry.

 

[ku1005]

actually, I made this more complex then need be.

I think all you have to do is show that the sequence is decreasing, and then solving for the inequality, using whole number intergers, n, by trial and error, must be greater then or equal to 5.

ie N = 5 so for all values of n>N bla bla etc...cheers

 

[pka]

“Using properties of limits, show that there exists an N in the set of Positive Integers such that” The key here is the phrase “Using properties of limits”
Note that \(\displaystyle \L\frac{{\left( {n + 1} \right)^2 }}{{2n^2 }} \to \frac{1}{2}\).
Then use the definition of sequence convergence: \(\displaystyle \L\varepsilon = \frac{1}{4}\) then we know that \(\displaystyle \L\left( {\exists N} \right)\left[ {n \ge N} \right] \Rightarrow \left| {\frac{{\left( {n + 1} \right)^2 }}{{2n^2 }} - \frac{1}{2}} \right| < \varepsilon = \frac{1}{4}\).
From that we get \(\displaystyle \L- \frac{1}{4} < \frac{{\left( {n + 1} \right)^2 }}{{2n^2 }} - \frac{1}{2} < \frac{1}{4}\; \Rightarrow \quad \frac{{\left( {n + 1} \right)^2 }}{{2n^2 }} < \frac{3}{4}\)

 

[ku1005]

i see!!!!....thanks for that...i was trying to do something along those lines...it just didnt hapin lol, so thanks.