Question #86545

The limit is #0#.

It makes sense because both #1/n# and #(-1)/n# go to #0# as #nrarroo#.

In fact, you can prove that

#lim_(nrarroo) (-1)^n/n =0#

using
#-1/n <= (-1)^n/n <= 1/n#

and the discrete version of the squeeze theorem (aka sandwich or pinch theorem).

For #lim_(ntooo)(-1)^n/n#, it converges at #0#.

To determine this limit, the best way to do so is using the idea of the squeeze theorem. Similar in Calculus I for limits of functions, this idea states that as long as #a_n<=b_n<=c_n#, the limits of two sequences, #a_n# and #c_n#, can both equal some #L# value, indicating that the limit of #b_n# must equal #L# as well.

This means that as the two sequences converge to some number or diverge, the sequence in the middle (the original sequence) must also do the same since it is between the two as they "squeeze" to the same final result. Here is a good example online:

http://supportcentre.maths.nuim.ie/documents/uploads/user/SqueezeTheoremforSequences.pdf

For alternating sequences like #(-1)^n/n#, there is a special case of the squeeze theorem: the absolute value theorem.

If you take the absolute value of a sequence #a_n# and #lim_(ntooo)|a_n| = 0#, then the limit of the original sequence must converge to #0# as long as #-|a_n|<=a_n<=|a_n|#.

Let's begin by taking the absolute value of the sequence we are solving for:

#lim_(ntooo)|a_n|=lim_(ntooo)abs((-1)^n/n)=lim_(ntooo)(1)^n/n=lim_(ntooo)1/n=0#

We can assume #1^n# goes to #1# since the sequence will be just #1, 1, 1, 1,...# Let's check with the squeeze theorem:

#-1/n<=(-1)^n/n<=1/n=>#true!

#lim_(ntooo)(-1/n)=lim_(ntooo)(1/n)=L=0#

So, #lim_(ntooo)(-1)^n/n=0# by squeeze theorem!

Since this is always the case for alternating sequences, you can just show the absolute sequence approaching to zero and state what theorem you used (squeeze theorem or absolute value theorem).