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Does divergence of $\sum a_k$ imply divergence of $\sum \frac{a_k}{1+a_k}$?

Note: $a_k > 0 $

I understand that looking at the contrapositive statement, we can say that the convergence of the latter sum implies $\frac{a_k}{1+a_k}\rightarrow 0$ but from here is it possible to deduce that $a_k\rightarrow 0$ because it is not completely straightforward. If we assume $a_k$ to be convergent, this trivially follows but it could diverge in which case this is nontrivial to me.

Aakash Lakshmanan
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3 Answers3

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If $a_k\geq 0$ for every $k$, then you can reason as follows.

If $(a_k)$ does not converge to $0$, then there exists a subsequence $(a_{n_k})$ with limit $l\in (0,+\infty]$.

It is then easy to verify that $$ \frac{a_{n_k}}{1+a_{n_k}} \to \begin{cases} 1, & \text{if}\ l=+\infty,\\ \frac{l}{1+l}\neq 0, &\text{if}\ l\in (0,+\infty), \end{cases} $$ hence $\sum \frac{a_k}{1+a_k}$ cannot converge, so that diverges to $+\infty$.

Consider now the case $\lim_k a_k = 0$. Then there exists $N\in\mathbb{N}$ such that $0\leq a_k \leq 1$ for every $k\geq N$, so that $$ \frac{a_k}{1+a_k} \geq a_k/2 \qquad \forall k\geq N, $$ and again the series diverges by comparison.

Remark: the point is that $$ \frac{a_k}{1+a_k} \geq \begin{cases} 1/2, &\text{if}\ a_k > 1,\\ a_k / 2, & \text{if}\ a_k\in [0,1]. \end{cases} $$

Rigel
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  • what is $a_n$ does not have a limit at all – Guy Fsone Dec 20 '17 at 09:13
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    That's the first case (the sequence does not converge or converge to a number different from $0$). – Rigel Dec 20 '17 at 09:14
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    What if $a_n$ just oscillates without a limit? It simply diverges? – Aakash Lakshmanan Dec 20 '17 at 09:22
  • @AakashLakshmanan that was actually the point of my question – Guy Fsone Dec 20 '17 at 09:24
  • I was rephrasing to make it more clear to Rigel as he only considered the infinite case – Aakash Lakshmanan Dec 20 '17 at 09:25
  • Even if $(a_n)$ oscillates without a limit, it always admits a subsequence that converges or diverges to $+\infty$ (take, e.g., the subsequence that realizes the $\limsup$). – Rigel Dec 20 '17 at 09:28
  • Nevermind, I understand now! Bolzano–Weierstrass theorem! – Aakash Lakshmanan Dec 20 '17 at 09:28
  • If the subsequence $a_{n_k} / (1+a_{n_k})$ converges to a number different from $0$, then the whole sequence $a_k/(1+a_k)$ cannot converge to $0$. – Rigel Dec 20 '17 at 09:36
  • @AakashLakshmanan It does not have to do anything with Bolzano-Weierstrass theorem, what Rigel uses is that $\limsup$ always exist, it is either finite or infinite and both cases is elaborated nicely. – Shashi Dec 20 '17 at 09:44
  • @Rigel what is your sequence behave like $\sin n$? the limit does not exists at all neither finite nor infinte? – Guy Fsone Dec 20 '17 at 12:40
  • Nevertheless the set of the limits of its subsequences is $[-1,1]$. In particular, there exists a subsequence $(n_k)$ such that $\sin n_k \to 1$ as $k\to +\infty$. – Rigel Dec 20 '17 at 17:08
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If $\sum_n \frac{a_n}{1+a_n}$ converges, $\frac{a_n}{1+a_n}$ goes to $0$ as $n$ goes to $\infty$, hence $a_n$ goes to $0$ as $n\to \infty$.

But since $\lim_{n\to \infty} \frac{a_n}{\frac{a_n}{1+a_n}} = 1$, for large enough $n$, $\left|\frac{a_n}{\frac{a_n}{1+a_n}} - 1 \right|\leq \frac 12$, hence $a_n\leq \frac 32 \frac{a_n}{1+a_n}$. Therefore, $\sum a_n$ converges by comparison.

Gabriel Romon
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In the case where the $a_n$ can change sign, let $$b_n = \frac{(-1)^n}{\sqrt{n}}\quad \text{and}\quad a_n=\frac{b_n}{1-b_n} = \frac{(-1)^n}{\sqrt{n}}\frac{1}{1-\frac{(-1)^n}{\sqrt{n}}} = \frac{(-1)^n}{\sqrt{n}} + \frac{1}{n} + O\left(\frac{1}{n^{3/2}}\right)$$ then one has $b_n = \frac{a_n}{1+a_n}$, the series $\sum a_n$ diverges but the series $\sum b_n$ converges.

Gribouillis
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