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If $\sum_{n=0}^\infty\frac{a_n}{b_n}$ converges and $\sum_{n=0}^\infty\frac{a_n^2}{b_n^2}$ converges, where $(a_n + b_n)b_n \ne 0$ for every $n \in \mathbb{N}$ , then show that $\sum_{n=0}^\infty \frac{a_n}{a_n + b_n}$ also converges.

I have proved it for non-negative $a_n$ and $b_n$, but I am unable to do it for the other cases. Any help will be greatly appreciated.

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You should learn how to write in LaTeX! – Mercy King Sep 1 '12 at 13:05
Why do you need the condition $(a_n+b_n)b_n \ne 0$ for? – Mercy King Sep 1 '12 at 13:16
To have all the sequences well-defined . – Ester Sep 1 '12 at 13:18
Isn't $a_n+b_n \ne 0$ enough?! – Mercy King Sep 1 '12 at 13:28
What about the other 2 sequences? – Ester Sep 1 '12 at 13:31
up vote 5 down vote accepted

Considering $c_n=\frac{a_n}{b_n}$, one assumes that $C=\sum\limits_nc_n$ and $C'=\sum\limits_nc_n^2$ both converge and that $c_n\ne-1$ for every $n$, and one wants to show that $D=\sum\limits_nd_n$ converges, with $d_n=\frac{c_n}{1+c_n}$.

  • Since $d_n=c_n-e_n$ with $e_n=\frac{c_n^2}{1+c_n}$ and $C$ converges, $D$ converges if and only if $E=\sum\limits_ne_n$ does.
  • Since $C$ converges, $c_n\to0$ hence $|c_n|\leqslant\frac12$ for every $n$ large enough, and then $|1+c_n|\geqslant\frac12$.
  • For every $n$ large enough, $\left|e_n\right|\leqslant2c_n^2$.
  • Hence the convergence of $C'$ implies the (absolute) convergence of $E$. Done.
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Note: This proof is for the series $\sum_n A_n$ and $\sum_n B_n$ with positive terms.
Given $ \sum_{n=0}^\infty\frac{a_n}{b_n} $ and $ \sum_{n=0}^\infty\frac{a^2_n}{b^2_n} $ converge, then this implies $ \sum_{n=0}^\infty \left( \frac{a_n}{b_n} + \frac{a_n^2}{b^2_n} \right ) $ converges.

Let $A_n = \frac{a_n}{a_n+b_n} $ and $ B_n = \frac{a_n}{b_n} + \frac{a_n^2}{b^2_n} $, then we have

$$ \lim_{ n \rightarrow \infty } \frac{A_n}{B_n}=1 \,.$$

Use the fact that $ \lim_{ n \rightarrow \infty } \frac{a_n}{b_n} = 0 $ (since the series $ \sum_{n=0}^\infty\frac{a_n}{b_n} $ is a convergent series) to prove the above limit.

Now, since $\lim_{ n \rightarrow \infty } \frac{A_n}{B_n}=1$, then by the limit comparison test, the two series converge since $\sum_{n=0}^{\infty} B_n $ converges.

limit of $\frac{A_n}{B_n}$

$$ \frac{A_n}{B_n} = \frac{\frac{a_n}{a_n+b_n}}{\left( \frac{a_n}{b_n} + \frac{a_n^2}{b^2_n} \right )} = \frac{1}{\left(\frac{a_n}{b_n}+1\right)^2} \rightarrow 1 $$ as $n \rightarrow \infty$.

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You can only do this for positive sequences A(n) and B(n) which may not be the case here – Ester Sep 1 '12 at 14:43
Example of the problem mentioned by @Ricky: $B_n=(-1)^n/\sqrt{n}$, $A_n=B_n+1/n$. – Did Sep 1 '12 at 19:36
About the Note at the beginning of this post, if $B_n\gt0$ for every $n$ and $\sum\limits_nB_n$ converges, then $c_n=a_n/b_n$ is positive for $n$ large enough. Then $d_n=a_n/(a_n+b_n)=c_n/(1+c_n)$ hence $0\lt d_n\lt c_n$ and the convergence of the series $\sum\limits_nB_n$ alone implies the convergence of $\sum\limits_nd_n$. In other words, the case when $B_n\gt0$ for every $n$ is much easier than the question asked. – Did Sep 2 '12 at 7:38
@SerialDownvoter:What's the downvote for? – Mhenni Benghorbal Sep 9 '12 at 17:04
@did:It is a solution for a part of the question. – Mhenni Benghorbal Sep 9 '12 at 17:05

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