# Consider the sequence a_n = ((-1)^n)/(sqrt(n)) What is the limit as n approaches infinity of a_n ? We can use the fact that if `lim_(n->oo)|a_n|=0` then `lim_(n->oo)a_n=0`

(We can use the squeeze theorem; since `-|a_n|<=a_n<=|a_n| ` the result holds.)

So `lim_(n->oo)|((-1)^n)/sqrt(n)|=lim_(n->oo)1/sqrt(n)` . For `n>=1` this function is monotonically decreasing and is bounded above by 1. A sequence that is bounded and monotonic converges. (Use the fact that...

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

We can use the fact that if `lim_(n->oo)|a_n|=0` then `lim_(n->oo)a_n=0`

(We can use the squeeze theorem; since `-|a_n|<=a_n<=|a_n| ` the result holds.)

So `lim_(n->oo)|((-1)^n)/sqrt(n)|=lim_(n->oo)1/sqrt(n)` . For `n>=1` this function is monotonically decreasing and is bounded above by 1. A sequence that is bounded and monotonic converges. (Use the fact that `lim_(x->oo)1/x^(r)=0` for r>0)

-------------------------------

Therefore `lim_(n->oo)((-1)^n)/sqrt(n)=0`

--------------------------------

Proof: ` ``lim_(n->oo)1/sqrt(n)=0` means that for any `epsilon>0` there exists M>0 such that n>M implies that `|1/sqrt(n)-0|<epsilon`

Let `M=1/(epsilon^2)` Then `n>M==> n>1/epsilon^2`

`sqrt(n)>1/epsilon`

`1/sqrt(n)<epsilon` as required.

Approved by eNotes Editorial Team