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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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`


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.

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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.


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