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

  • 5
    $\begingroup$ 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. $\endgroup$ – Thomas Andrews Feb 6 '13 at 17:43
  • $\begingroup$ @jonas-meyer hi .i use comparison test but its doesn't help me why second condition is redundant? $\endgroup$ – M.H Feb 6 '13 at 17:49
  • 4
    $\begingroup$ @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. $\endgroup$ – Julián Aguirre Feb 6 '13 at 17:53
  • $\begingroup$ @Julián: Thanks for the correction. I'll delete the wrong comment. $\endgroup$ – Jonas Meyer Feb 6 '13 at 18:33

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.

  • $\begingroup$ Nicely done for those who want to avoid Taylor expansions. +1 $\endgroup$ – Julien 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 $$ \frac{c_n}{1+c_n}=c_n(1-c_n+O(c_n^2))=c_n-c_n^2+O(c_n^3)=c_n+O(c_n^2). $$

  • 1
    $\begingroup$ 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. $\endgroup$ – robjohn Feb 6 '13 at 18:20
  • 1
    $\begingroup$ @robjohn One hypothesis is that $\sum c_n^2$ converges... $\endgroup$ – Julien Feb 6 '13 at 18:29
  • $\begingroup$ 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. $\endgroup$ – robjohn Feb 6 '13 at 18:43
  • $\begingroup$ @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... $\endgroup$ – Julien 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.

  • $\begingroup$ Do you mean: a comparison theorem which would work for general terms that are not nonnegative...? $\endgroup$ – Julien Feb 6 '13 at 18:31
  • 1
    $\begingroup$ @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 $0\le B_n-A_n\le C_n-A_n$. $\endgroup$ – Julián Aguirre Feb 7 '13 at 10:19
  • $\begingroup$ Thank you, Julian, +1. Now I wonder how I have spent all these years without ever using this extended criterion... $\endgroup$ – Julien 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. $c_n = (-1)^n/n)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.