Convergent or divergent, how do you do a problem as such.

Question

astrophysics · Answer

$$a_n = \frac{ n^3 }{ n }$$

@ganeshie8

astrophysics · Answer

$$a_n=\frac{ n^3 }{ n! }$$ should be this.

astrophysics · Answer

$$a_n=\frac{ n*n*n }{n!  }$$ not really sure about the n!

loser66 · Answer

converge to 0

loser66 · Answer

Let expand the denominator, it is n(n-1)(n-2)(n-3)........
just take the first 4 terms and open the bracket, you get n^4 +....., right? n^3/n^4 = 1/n 
so that the leading term of the fraction is 1/n , and it limit is 0 when n goes to infinitive.

anonymous · Answer

Assuming this is a sequence: To show that a sequence converges, you must establish that the sequence is monotonically increasing and bounded above, or decreasing and bounded below.

Assuming this is a series: Use the ratio test. It works well with factorial expressions.

astrophysics · Answer

Sorry but, why is n! = n(n-1)(n-2)(n-3)...I don't remember too much about sequences.