a_n = (-1)^n(n/(n+1)) Determine the convergence or divergence of the sequence with the given n'th term. If the sequence converges, find its limit.

A sequence of real numbers (a_n) is said to be convergent if forall epsilon>0, there exists N in NN such that forall n>N  then |a_n-a|If the sequence is convergent, then a is called the limit of the sequence.
In other words the sequence is convergent if the terms tend to a single value a n increases to infinity. That single value is called the limit of the sequence.
Let us first calculate limit of sequence with nth term n/(n+1).
lim_(n to infty)n/(n+1)=
Divide both numerator and the denominator by n.
lim_(n to infty)(n/n)/(n/n+1/n)=lim_(n to infty)1/(1+1/n)=
Since lim_(n to infty)1/n=0 we have
lim_(n to infty)1/(1+1/n)=1/1=1
This part of the sequence converges to 1 however, (-1)^n has two distinct values -1 for odd n and 1 for even n (these types of sequences are called alternating sequences). Therefore, the sequence will have two distinct accumulation points 1 and -1. Therefore, if we choose epsilon<2 and either of the two points e.g. 1 , we can always find some term for which |1-a_n|>2 no matter how big the N we choose.
Therefore, we conclude that the sequence is divergent.
The image below shows first 50 terms of the sequence. We can clearly see the two accumulation points 1 and -1. 
https://en.wikipedia.org/wiki/Limit_of_a_sequence

https://en.wikipedia.org/wiki/Limit_point