# $a_j \geq 0, \sum_{j=1}^{\infty} a_j$ divergent $\implies \sum_{j=1}^{\infty}\frac{a_j}{1 + a_j}$

Suppose that $a_j \geq 0$ and that the $\sum_{j=1}^{\infty} a_j$ diverges. Prove that the following series diverges:

$$\sum_{j=1}^{\infty}\frac{a_j}{1 + a_j}$$ Hint: first show that if it converges, then $a_j$ converges to $0$.

Thanks!

The graph of $f(x)=x/(1+x)$ is increasing on $[0,\infty)$ with range $[0,1)$ (an asymptote as $x \to \infty$). So if we assume that $\sum a_j/(1+a_j)$ converges, its $j^{th}$ term approaches zero, so that from the remarks about $f(x)$ we get $a_j \to 0.$ We may then choose some $N$ so that for $j>N$ we have $a_j \le 1,$ so that also $a_j^2 \le a_j$ for those $j$.
But then $$a_j = \frac{a_j}{1+a_j}+\frac{a_j^2}{1+a_j} \le 2 \frac{a_j}{1+a_j}.$$ This is incompatible with the divergence of $\sum a_j.$
Hint (slightly different approach): if $a_j > 1$, we can write $$\frac{a_j}{1 + a_j}>\frac{a_j}{a_j + a_j} = \frac 12$$ If on the other hand $a_j \leq 1$, we can write $$\frac{a_j}{1 + a_j}>\frac{a_j}{1 + 1} = \frac{a_j}2$$
• @mflowww right. Consider the following division: either there are infinitely many $j$ for which $a_j \geq 1$, or there are finitely many, which means that there is some $J$ for which $a_j \leq 1$ for $j>J$. You can prove divergence in each case; in the first, we note that $\frac{a_j}{1 + a_j}$ doesn't converge to zero, which means that the series can't converge. In the second, we can make a comparison with $\sum a_j$. – Omnomnomnom Nov 18 '13 at 2:22