`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...

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|<epsilon.`

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 `n`th term `n/(n+1).`

`lim_(n to infty)n/(n+1)=`

Divide...

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|<epsilon.`

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 `n`th 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.`