First it' not enough to show that `lim_(n->infty)a_n ne0`. For example series `sum1/n` is the case in which `lim_(n->infty)1/n=0` while `lim_(n->infty)sum_(n=1)^(infty)1/n=infty`

**Meaning of** `Theta.`

`f(x)=Theta(g(x))` means that `c_1g(x)leq f(x)leq c_2g(x)` for some`c_1,c_2 in RR`. To put it more simply it means that function `f` is behaving (growing and falling) at approximately the same rate (if we forget about constants) as function `g.`

For example: `3x^2=Theta(x^2)` because `4x^2 leq 3x^2 leq 2x^2`

`15e^x-x^2+2=Theta(e^x)`

For more on asymptotic notation see e.g. big O or consult any the **literature**:* Knuth, Donald E.: The Art of Computer Programming vol.1* or *Cormen, Leiserson, Rivest, Stein: Introduction to Algorithms*

**3 cases:**

Case 1 - What hapens if `a_n` is divergent e.g. `a_n=n`

Case 2 - What hapens if `a_n` is convergent and is falling (or growing if negative) at the same rate as `1/n^p` for `p leq1`

Case 3 - This case is only here to fill up what happens if `p >1` but it does not actually interest us because in that case our series `sum a_n` is convergent what is in contradiction with our assumption.

I really hope this helpes.

We will use necessary condition for series to be convergent.

For series `sum a_n` to be convergent the following condition must be met: `lim_(n->infty)a_n=0`.

Let's apply that to our series.

**case 1**

First assume `lim_(n->infty)ncdota_n >0` (case `lim_(n->infty)ncdota_n <0` is equivalent) and `a_n` divergent.

`lim_(n->infty)a_n/(1+a_n)=` (devide both numerator and denominator by `a_n`)

`lim_(n->infty)1/(1/a_n+1)=1` because `1/a_n->0` because `a_n` is divergent.

**case 2**

If `a_n` not divergent then `a_n=Theta(1/n^p)` for `p leq1`, but we know that `sumTheta(1/n^p)` is divergent series (which can be proven by using integral test for convergence).

**case 3**

If `lim_(n->infty)ncdota_n=0=>a_n=Theta(1/n^p)` for `p>0` which would mean that our series is convergent which is not.