# `a_n = (1+(-1)^n)/n^2` Determine the convergence or divergence of the sequence with the given n'th term. If the sequence converges, find its limit.

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**Method I**

Break the `n`th term into two separate fractions

`lim_(n to infty)(1/n^2+(-1)^n/n^2)=`

`1/n^2` tends to infinity and `(-1)^n/n^2` is equal to `-1/n^2` for odd `n` and `1/n^2` for even `n` and both those expressions tent to infinity as `n` goes to infinity. Therefore we get

`0+0=0`

**Sequence is convergent and its limit is equal to 0.**

**Method II**

`lim_(n to infty)a_n=lim_(n to infty)(1+(-1)^n)/n^2`

Let us break this into two cases (one for even and one for odd `n`). If both cases give the same result then the sequence has a single accumulation point and is thus convergent.

`n=2k,` `k in NN` **(n is even)**

`lim_(n to infty)(1+(-1)^(2k))/n^2=lim_(n to infty)(1+1)/n^2=lim_(n to infty)2/n^2=0`

`n=2k-1,` `k in NN` **(n is odd)**

`lim_(n to infty)(1+(-1)^(2k-1))/n^2=lim_(n to infty)(1-1)/n^2=lim_(n to infty)0/n^2=0`

**Both limits are equal to zero hence, the sequence is convergent and its limit is equal to zero.**

The image below shows the first 20 terms of the sequence. We can see that even-numbered terms converge to zero while odd-numbered terms forms a stationary subsequence (it is always equal to zero).