someone can solve this !

[kizim]

a_n=[n/2] _{- (}n²/2n+1)

i want to know if a_n converges to some limit .. if yes i want to find it and explain ????

 

[Deleted member 4993]

a_n=[n/2] _{- (}n²/2n+1)

i want to know if a_n converges to some limit .. if yes i want to find it and explain ????

is your problem:

\(\displaystyle a_n = \dfrac{n}{2} \ - \ \dfrac{n^2}{2n+1}\)

or

\(\displaystyle a_n = \ \dfrac{\frac{n}{2} \ - \ n^2}{2n+1}\)

or something else?

Please share your work with us.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/forum/th...217#post322217

We can help - we only help after you have shown your work - or ask a specific question (e.g. "are these correct?")

 

[DrPhil]

a_n=[n/2] _{- (}n²/2n+1)

i want to know if a_n converges to some limit .. if yes i want to find it and explain ????

Always hard to type rational expressions inline - if (2n+1) is the denominator of the second fraction, you need to have parentheses around the denominator.

a_n=[n/2] _- n²/(2n+1) --> \(\displaystyle \displaystyle a_n = \dfrac{n}{2} - \dfrac{n^2}{2n + 1} \)

Combine the two fractions over the common denominator, \(\displaystyle 2(2n + 1)\). Looks yo me like the \(\displaystyle n^2\) terms will cancel. Then you can take the limit as n becomes large.