# Thread: Divergent fractional series

1. ## Divergent fractional series

Given that $\displaystyle \sum{a_n}$ diverges, and $\displaystyle a_n\geq{0}$, prove $\displaystyle \sum{\frac{a_n}{a_n+1}$ diverges

So for any given $\displaystyle M\in{\mathbb{R}}, \exists N\in{\mathbb{N}}$ such that when $\displaystyle n\geq{N}$, $\displaystyle \sum^n_{k=1}{a_k}\geq{M}$

Where do we go from here?

2. ## Re: Divergent fractional series

Originally Posted by I-Think
Given that $\displaystyle \sum{a_n}$ diverges, and $\displaystyle a_n\geq{0}$, prove $\displaystyle \sum{\frac{a_n}{a_n+1}$ diverges
So for any given $\displaystyle M\in{\mathbb{R}}, \exists N\in{\mathbb{N}}$ such that when $\displaystyle n\geq{N}$, $\displaystyle \sum^n_{k=1}{a_k}\geq{M}$ Where do we go from here?
You have two cases: $\displaystyle (a_n)\not\to 0$ and $\displaystyle (a_n)\to 0$.

Consider the implications of each case.

3. ## Re: Divergent fractional series

If $\displaystyle \lim_{n \rightarrow \infty} a_{n} \ne 0$ , then also $\displaystyle \lim_{n \rightarrow \infty} \frac{a_{n}}{1+a_{n}} \ne 0$ and the series $\displaystyle \sum_{n} \frac{a_{n}}{1+a_{n}}$ diverges. If $\displaystyle \lim_{n \rightarrow \infty} a_{n} = 0$ , then for n 'large enough' is $\displaystyle \frac{a_{n}}{1+a_{n}}> \frac{a_{n}}{2}$ and, because the series $\displaystyle \sum_{n} \frac{a_{n}}{2}$ diverges, also the series $\displaystyle \sum_{n} \frac{a_{n}}{1+a_{n}}$ diverges...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

4. ## Re: Divergent fractional series

Okay, contemplated on hint, believe I got a solution

For $\displaystyle a_n\rightarrow{x\neq{0}}$, $\displaystyle \frac{a_n}{1+a_n}\rightarrow{y\neq{0}}$, so result follows

For $\displaystyle a_n\rightarrow{0}$, assume $\displaystyle \sum\frac{a_n}{1+a_n}$ convergent

For any given $\displaystyle n$, let $\displaystyle Z=max(1+a_i)$, $\displaystyle 1\leq{i}\leq{n}$

Choose $\displaystyle M$ so $\displaystyle \frac{M}{Z}>LUB(\sum\frac{a_n}{1+a_n})$, choose $\displaystyle N$ so that for $\displaystyle n\geq{N}$, $\displaystyle \sum{a_n}>M$

Now $\displaystyle \sum{a_n}<Z\sum\frac{a_n}{1+a_n}<M$

But $\displaystyle \sum{a_n}>M$, so a contradiction has occurred. The result follows

Is this proof satisfactory?

5. ## Re: Divergent fractional series

Just saw chisigma's proof, it's slicker than mine, but I believe this is still correct, no?