# how prove $\sum_{n=1}^\infty\frac{a_n}{b_n+a_n}$is convergent?

Let$a_n,b_n\in\mathbb R$ and $(a_n+b_n)b_n\neq 0\quad \forall n\in \mathbb{N}$. The series $\sum_{n=1}^\infty\frac{a_n}{b_n}$ and $\sum_{n=1}^\infty(\frac{a_n}{b_n})^2$ are convergent.

How to prove that$$\sum_{n=1}^\infty\frac{a_n}{b_n+a_n}$$ is convergent.

Thanks in advance

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Note, you can rewrite this letting $c_n=\frac{a_n}{b_n}$. Then $\sum c_n$ converges as does $\sum c_n^2$, and you want to show that $\sum \frac{c_n}{1+c_n}$ converges. – Thomas Andrews Feb 6 '13 at 17:43
@jonas-meyer hi .i use comparison test but its doesn't help me why second condition is redundant? – Maisam Hedyelloo Feb 6 '13 at 17:49
@JonasMeyer There is no assumption about the sign of $a_n/b_n$. The condition $\sum(a_n/b_n)^2<\infty$ is not redundant. – Julián Aguirre Feb 6 '13 at 17:53
@Julián: Thanks for the correction. I'll delete the wrong comment. – Jonas Meyer Feb 6 '13 at 18:33

## 4 Answers

Note that $$\frac{a_n}{b_n}-\frac{a_n}{a_n+b_n}=\frac{a_n^2}{a_nb_n+b_n^2}\tag{1}$$ Since $\sum\frac{a_n}{b_n}$ converges, there exists an $N$ so that for $n>N$, we have $$\left|\frac{a_nb_n}{b_n^2}\right|=\left|\frac{a_n}{b_n}\right|<\frac12\tag{2}$$ Then for $n>N$ we also have that $$0\le\frac{a_n^2}{a_nb_n+b_n^2}\le2\frac{a_n^2}{b_n^2}\tag{3}$$ So by comparison to $\sum2\frac{a_n^2}{b_n^2}$, we have that $\sum\frac{a_n^2}{a_nb_n+b_n^2}$ converges. Then using $(1)$ we have that $$\sum\frac{a_n}{a_n+b_n}=\sum\frac{a_n}{b_n}-\frac{a_n^2}{a_nb_n+b_n^2}\tag{4}$$ also converges.

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Nicely done for those who want to avoid Taylor expansions. +1 – 1015 Feb 6 '13 at 18:55

Hint: Following Thomas, let us denote $c_n=a_n/b_n$. Since $\sum c_n$ converges, $c_n$ tends to $0$ and $$\frac{c_n}{1+c_n}=c_n(1-c_n+O(c_n^2))=c_n-c_n^2+O(c_n^3)=c_n+O(c_n^2).$$

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Why is $O(c_n^2)$ convergent? let $c_n=\frac{(-1)^n}{\log(n+1)}$, then $\sum\limits_{n=1}^\infty c_n$ converges, but $\sum\limits_{n=1}^\infty c_n^2$ does not. – robjohn Feb 6 '13 at 18:20
@robjohn One hypothesis is that $\sum c_n^2$ converges... – 1015 Feb 6 '13 at 18:29
Sorry, I missed that this was a hint (+1). There are several key points that need to be addressed for a full answer, and that was why I was picking nits. – robjohn Feb 6 '13 at 18:43
@robjohn Thanks! Given that $\sum c_n$ and $\sum c_n^2$ are assumed to converge, there are not so many key points to add to my hint to get a full answer... – 1015 Feb 6 '13 at 18:49

As suggested by Thomas let $c_n=a_n/b_n$. Then $\sum c_n$ and $\sum c_n^2$ converge. Since $\sum c_n$ converges, we have $\lim_{n\to\infty}c_n=0$. We may assume without loss of generality that $|c_n|\le1/2$. It is easy to check that $$x-2\,x^2\le\frac{x}{1+x}\le x,\quad |x|\le1/2.$$ Then $$c_n-2\,c_n^2\le \frac{c_n}{1+c_n}\le c_n.$$ The convergence of $\sum c_n/(1+c_n)$ follows from the (generalized) comparison theorem.

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Do you mean: a comparison theorem which would work for general terms that are not nonnegative...? – 1015 Feb 6 '13 at 18:31
@julien Yes. If $A_n\le B_n\le C_n$ and $\sum A_n$, $\sum C_n$ converge, then $\sum B_n$ converges. It is a little known extension of the classical comparison theorem. The proof is easy: apply the classical comparison criterion to the inequality $0\le B_n-A_n\le C_n-A_n$. – Julián Aguirre Feb 7 '13 at 10:19
Thank you, Julian, +1. Now I wonder how I have spent all these years without ever using this extended criterion... – 1015 Feb 7 '13 at 14:30

Here's a thought: (I'll let $c_n = a_n / b_n$ following Thomas Andrew's suggestion)

Reductions: assume that $|\sum_n c_n|, \sum_n c_n^2 < 1$, as well as $|c_n| < 1$ for all $n$. Then apply the geometric series formula to the summand, termwise in $n$: $$\sum_n \frac{c_n}{1+c_n} = \sum_n \sum_{m=0}^{\infty} (-1)^m c_n^{m+1} =\sum_n c_n + \sum_n \sum_{m=1}^{\infty} (-1)^m c_n^{m+1}$$ If we admit the second inequality for the time being, the first term on the RHS is a finite number, and using Tonelli we can show the second term is an absolutely convergent sequence by comparing $\sum_n c_n^{m+1}$ with products of $\sum_n c_n^2$.

I'm skeptical that this is perfectly rigorous, especially because we're dealing with a potentially highly non-absolutely-convergent situation (e.g. $c_n = (-1)^n/n)$.

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