Divergent fractional series

**I-Think** · August 22nd 2011, 04:20 PM

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

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

Where do we go from here?

**Plato** · August 22nd 2011, 05:00 PM

Originally Posted by I-Think

Given that $\sum{a_n}$ diverges, and $a_n\geq{0}$ , prove $\sum{\frac{a_n}{a_n+1}$ diverges
So for any given $M\in{\mathbb{R}}, \exists N\in{\mathbb{N}}$ such that when $n\geq{N}$ , $\sum^n_{k=1}{a_k}\geq{M}$ Where do we go from here?

You have two cases: $(a_n)\not\to 0$ and $(a_n)\to 0$ .

Consider the implications of each case.

**chisigma** · August 23rd 2011, 12:43 PM

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

Kind regards

$\chi$ $\sigma$

**I-Think** · August 25th 2011, 02:58 AM

Okay, contemplated on hint, believe I got a solution

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

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

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

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

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

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

Is this proof satisfactory?

**I-Think** · August 25th 2011, 03:03 AM

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

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