`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...

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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...

` `

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).