Sequence: Show that | n/(2n+1) - 1/2 | < Epsilon

[ku1005]

Hey guys, for the following Q i am having some trouble, I am really new to this stuff as well, and my lecturer goes quite fast through it, I understand it, just not quite sure how to set an answer out....any help or tips for other Q's like this appreciated!

"Show that, if n is positive integer, then for any Epsilon>0 there is a positive integer N such that

| n/(2n+1) - 1/2 | < Epsilon, whenever n>N "

So let an ("a-sub-n", the sequence terms) ---> L (the limit of the sequence)

Also knowing that a(n+1) /an < 1 then, the sequence must be decreasing.

ie 2n+1 / 2n+3 < 1 for all n

lso note that if n>N then 1/n < 1/N

but im not sure what exactl to do!!! argg...really frustrating

again...any tips greatly appreciated!!

rhys

 

[daon]

Let \(\displaystyle \epsilon > 0\) and \(\displaystyle N=\frac{1}{\epsilon}\) ... Or in your case, the first natural number greater than 1/\(\displaystyle \epsilon\), \(\displaystyle \lceil\frac{1}{\epsilon} \rceil\) which we can say since N is not bounded.

Then
\(\displaystyle \L
n > N \,\, \Rightarrow \,\, n> \frac{1}{\epsilon} \,\, \Rightarrow \,\, \frac{1}{n} < \epsilon \,\, \Rightarrow*\)

\(\displaystyle \L \\ | \frac{1}{4n+2} |< \epsilon \,\, \Rightarrow \,\, |\frac{(2n+1)-2n}{2(2n+1)}| < \epsilon \,\, \,\, \Rightarrow \,\, |\frac{1}{2} - \frac{n}{2n+1}| < \epsilon \,\,\)

The starred(*) implication comes from the fact that 1/n > 1/(4n+2). So if 1/n is less than epsilon, so is 1/(4n+2). The absolute values were thrown in just for completeness (to look like the definition). They needn't be there since the manipulations were always positive.

My choice for N was not magic, either. In most of these problems it is best to try and combine the |a_n - L| into a fraction and to simplify it to something nice with a series of less-than or equals. From that "simple" or "nice" derivation likw I did from1/(4n+2) to 1/n, set it equal to epsilon and solve for n. Then set N equal to that result. Get it? If not, sorry, I've been up all night doing my own math problems and am kinda tired!

 

[ku1005]

Thanks heaps!! I appreciate your reply and time!