# How can be proved the following? If `sum a_n` is a divergent serie having positive terms, then: `sum a_n/(1+a_n)` is a divergent serie as well.

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### 4 Answers

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.

Sorry, in my response for case 2, the image does not appear. I meant:

"What's the meaning of:

an = THETA(1/n^p) "

SUM(an/1+an)

and in my response for case 1, the same occurs. When I said:

"it's stated that1"

I meant:

"it's stated that

lim (an/an+1) = 1 "

Thanks a lot Tiburtius for you detailed answer.

There are some points I don't understand:

1: In case 1, it's stated that1 but, what is the relationship between this and the assumption made for this case 1? And, moreover, wouldn't be enough to decide that

is divergent the fact its limit is not 0?

2: What distinguishes this case from 1 or 3? What's the meaning of ? I don't know that that "theta" symbol.

3: About case 3, again, the "theta" symbol makes I can't understand what it's stated there.

I would really appreciate if you could clarify these doubts.