SOLUTION: if {{{1/2 }}}{{{1/3 }}}{{{1/4,}}}{{{ 1/5 }}}... {{{1/(n+1)...}}} is an infinite sequence, then t_n= (A){{{1/n}}} n is all natural numbers (B){{{1/(n+2)}}} n is all positive integ

Algebra ->  Sequences-and-series -> SOLUTION: if {{{1/2 }}}{{{1/3 }}}{{{1/4,}}}{{{ 1/5 }}}... {{{1/(n+1)...}}} is an infinite sequence, then t_n= (A){{{1/n}}} n is all natural numbers (B){{{1/(n+2)}}} n is all positive integ      Log On


   

Question 1148527: if 1%2F2+1%2F3+1%2F4%2C+1%2F5+... 1%2F%28n%2B1%29... is an infinite sequence, then t_n=
(A)1%2Fn n is all natural numbers
(B)1%2F%28n%2B2%29 n is all positive integers
(C)1%2F%28n%2B1%29 n is all integers
(D)1%2F%28n%2B1%29 n > 0
(E) None of the above
Found 2 solutions by Edwin McCravy, ikleyn:
Answer by Edwin McCravy(20064) About Me  (Show Source):
You can put this solution on YOUR website!
It's not (A), for that would be  1%2F11%2F2+1%2F3+1%2F4%2C+1%2F5+... 1%2F%28n%29...

It's not (B), for that would be  1%2F3+1%2F4+1%2F5%2C+1%2F6+... 1%2F%28n%2B2%29...

It's not (C), for that would be a "two-way" sequence that goes "left and right":

...1%2F%28n%2B1%291%2F%28-5%291%2F%28-4%291%2F%28-3%291%2F%28-2%291%2F%28-1%29 1%2F01%2F11%2F21%2F31%2F41%2F5...1%2F%28n%2B1%29...

but that contains the undefined term "1/0". So that's not it. 

If the following were listed I would pick this:

matrix%281%2C7%2C1%2F%28n%2B1%29%2C+%22%22%2Cn%2Cis%2Call%2Cpositive%2C+integers%29

I would hesitate to pick (D) for it says nothing about n being and integer.

But I would also hesitate to pick (E)  

However, after thinking about it, I would pick (D) after all because you are
given that it is an infinite sequence, so it would not be necessary to say that
n is an integer.

Edwin

Answer by ikleyn(52879) About Me  (Show Source):
You can put this solution on YOUR website!
.

Typical provocation.


Keep yourself as far as you can from teachers who give such problems.