This is a modification of a problem in Rudin.

Let $(a_n)$ be a sequence of positive numbers (that is $a_n \geq 0)$. Then $\sum a_n$ converges iff $\sum \frac{a_n}{1+a_n}$ converges.

My attempt:

$\Rightarrow$ $$\frac{a_n}{1+a_n} \leq a_n$$ and this follows from comparison test.


Since $$a_n = \frac{a_n}{1+a_n} (1+a_n) = \frac{a_n}{1+a_n} + \frac{a_n^2}{1+a_n}$$

it suffices to show that $\sum \frac{a_n^2}{1+a_n}$ converges. For this, it suffices to show that $(a_n)$ is bounded, because then the result follows from the comparison test. Indeed, let $M$ be an upperbound. Then

$$\frac{a_n^2}{1+a_n} \leq \frac{Ma_n}{1+a_n}$$

We will prove that $a_n \to 0$, and this will prove the boundedness.

Let $\epsilon > 0$. Choose $N$ such that $\frac{a_n}{1+a_n} < \frac{\epsilon}{1+ \epsilon}$ for $n \geq N$, which is possible since $\frac{a_n}{1+a_n} \to 0$ since the series converges.

Then, $n \geq N$ implies that $a_n < \epsilon$ and the result follows.

Is this correct? Is there an easier way?


Your proof seems fine, maybe for the second implication, we can simply note that, since $a_n \to 0$ eventually $a_n<1$ and then $$\frac{a_n^2}{1+a_n}<\frac{a_n}{1+a_n}$$

As an easier way, assuming that $a_n \to 0$ (otherwise both don't converge), we can use limit comparison test and since

$$\frac{a_n}{\left(\frac{a_n}{1+a_n}\right)}=1+a_n \to 1$$

we conclude that $\sum a_n$ converges $\iff \sum \frac{a_n}{1+a_n}$ converges.

  • $\begingroup$ It is not given that $a_n \to 0$ $\endgroup$ – user370967 Jul 25 '18 at 21:17
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    $\begingroup$ @Math_QED If $a_n \not \to 0$ both do not converge. $\endgroup$ – user Jul 25 '18 at 21:18
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    $\begingroup$ Yes, that might be easier. But I still don't see how $\sum a_n/(1+a_n) < \infty$ implies that $a_n \to 0$ (Is the proof I provided in my attempt correct?). In your response, you said "assuming $a_n \to 0$", but this is what has to be proven! $\endgroup$ – user370967 Jul 25 '18 at 21:32
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    $\begingroup$ (1) How exactly does it follow? Is it because $a_n/(1+a_n) \to 0 \implies a_n \to 0$? (2) If yes, is my proof of this correct (see my attempt)? $\endgroup$ – user370967 Jul 25 '18 at 21:35
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    $\begingroup$ If $b_n = \frac{a_n}{1+a_n}$ then $a_n = \frac{b_n}{1-b_n}$; so $b_n \to 0$ implies $a_n \to 0$. $\endgroup$ – Daniel Schepler Jul 25 '18 at 21:57

Claim. If $\sum_{n=1}^{\infty} \frac{a_n}{1+a_n}$ converges then $a_n\to 0$.

[If I understood correctly, this was the only subtle point. The rest of your proof was perfectly clear].

Proof of the claim. If $\sum_{n=1}^{\infty} \frac{a_n}{1+a_n}$ converges, then $\sum_{n=1}^{\infty} \left(\frac{1+a_n}{1+a_n}-\frac{1}{1+a_n}\right)$ converges, so $\sum_{n=1}^{\infty} \left(1-\frac{1}{1+a_n}\right)$ converges. By the $n$-th term test, the sequence $1-\frac{1}{1+a_n}$ must converge to $0$. Thus, $\frac{1}{1+a_n}\to 1$ as $n\to\infty$. From here it is easy to see that $1+a_n \to 1$ and so $a_n\to 0$ as $n\to\infty$.


Assume $\sum_{n=1}^\infty \frac{a_n}{1+a_n}$ converges.

We claim that $a_n \to 0$. Let $0 < \varepsilon < 1$ and pick $n_0 \in\mathbb{N}$ such that $n \ge n_0 \implies a_n < \frac{\varepsilon}{1-\varepsilon}$. For such $n$ we have

$$a_n < \frac{\varepsilon}{1-\varepsilon} \iff a_n(1-\varepsilon) < \varepsilon \iff a_n < \varepsilon (1+a_n) \iff \frac{a_n}{1+a_n} < \varepsilon$$

Now pick $M > 0$ such that $1+a_n \le M, \forall n\in\mathbb{N}$.

We have

$$\sum_{n=1}^\infty a_n \le \sum_{n=1}^\infty \frac{Ma_n}{1+a_n}$$

which converges so $\sum_{n=1}^\infty a_n$ also converges by the comparison test.

  • $\begingroup$ Thanks. I think there is an issue in your proof if $\epsilon \geq 1$. $\endgroup$ – user370967 Jul 25 '18 at 22:05
  • $\begingroup$ @Math_QED True, just assume $0 < \varepsilon < 1$ so that $\frac{\varepsilon}{1-\varepsilon} > 0$. $\endgroup$ – mechanodroid Jul 25 '18 at 22:11

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