# Divergence of $\sum \frac{a_j}{1+a_j}$

Given: $a_j >0$ and $\sum a_j$ diverges. Show that $\sum \frac{a_j}{1+a_j}$ diverges.

Hint: show that if it converged, $a_j$ -> 0.

I don't understand how to think about this problem. Is there a convergence test I should use? I tried starting with the hint, but don't know how to conclude anything about the individual parts from the fraction's supposed convergence.

Following the hint: If $a_j\gt 0$ has limit $0$, then for large enough $j$, we have $0 \lt a_j\le 1$. Thus for large enough $j$, we have $\frac{a_j}{a_j+1}\ge \frac{a_j}{2}$. Thus by Comparison $\sum_{j=1}^\infty \frac{a_j}{a_j+1}$ diverges.

Added: We prove a detail that was left out in the hint. Suppose that $\sum \frac{a_j}{a+a_j}$ converges. Then $\lim_{j\to\infty} a_j=0$. For suppose this is not the case. Then there is an $\epsilon\gt 0$ such that $a_j\gt \epsilon$ for arbitrarily large $j$. If for a particular $j$, we have $a_j\ge 1$, then $\frac{a_j}{1+a_j}\ge \frac{1}{2}$. If $\epsilon \lt a_j\lt 1$, then $\frac{a_j}{1+a_j} \gt \frac{\epsilon}{2}$.

• But how can we show that $a_j$ has limit 0 if the series converges? – kiwifruit Apr 19 '14 at 3:56
• It is a standard and important theorem. Informally, if the limit of the $a_i$ is not $0$, the partial sums $\sum_1^n a_i$ bounce around, don't settle down. It is not hard to write an "$\epsilon$-$N$" proof. Note that the converse need not hold. We can have $\lim a_i=0$ but divergence of the series. Example{ $1=\frac{1}{2}+\frac{1}{3}+\cdots$. – André Nicolas Apr 19 '14 at 4:03
• The theorem is the following. Suppose that $\sum_{i=1}^\infty c_i$ converges. Then $\lim_{i\to\infty}c_i=0$. Absolutely general. Equivalently, if $\lim_{i\to\infty} c_i$ does not exist, or exists but is $\ne 0$, then $\sum_1\infty c_i$ diverges. – André Nicolas Apr 19 '14 at 4:21
• I have added a working out of $a_j\to 0$ from the hypothesis that $\frac{a_j}{1+a_j}\to 0$. – André Nicolas Apr 19 '14 at 4:35
• Soeey, was away. (1) We suppose that $\sum \frac{a_i}{1+a_i}$ converges. (2) We show that this implies that the $a_i\to 0$. (3) We conclude that after a while our terms are $\ge \frac{a_i}{2}$. (4) Since $\sum a_i$ diverges, this implies our series does. No circularity. Slightly convoluted chain of reasoning, but correct. It was the suggested path, and it works. – André Nicolas Apr 19 '14 at 19:48

Hint: To show the series diverges, notice that we have inequality $$\frac{a_{j}}{1+a_{j}}\geq\min\left(\frac{1}{2},\frac{a_{j}}{2}\right).$$ This inequality follows since either $a_j\geq\frac{1}{2}$, in which case $\frac{a_j}{1+a_j}\geq \frac{1}{2}$ or $0<a_j\leq \frac{1}{2}$, in which case $\frac{a_j}{1+a_j}\geq\frac{a_j}{2}$.

Hint: Since $1 + a_j \to 1$, we have that $1 + a_j < 2$ eventually. Then ponder the fact that

$$\frac{a_j}{1 + a_j} > \frac{a_j}{2}$$

• But how can we use the fact that $a_j -> 0$ to prove divergence, if by the hint this only holds when the series converges? – kiwifruit Apr 19 '14 at 16:08