# Suppose $a_n \geq 0$, and $\sum a_n$ diverges, and $\lim a_n = 0$. Show that $\sum \frac{a_n}{1+a_n}$ diverges.

Can someone please verify my proof sketch?

Suppose $a_n \geq 0$, and $\sum a_n$ diverges, and $\lim a_n = 0$. Show that $\displaystyle{\sum \frac{a_n}{1+a_n}}$ diverges.

Let $\epsilon > 0$. Then,

\begin{eqnarray} \exists N \in \mathbb{N} \ \ s.t. \ \ \forall n > N, 0 \leq a_n < \epsilon \end{eqnarray}

This implies that \begin{eqnarray} \forall n > N, \displaystyle{\frac{a_n}{1+\epsilon} < \frac{a_n}{1+a_n}} \end{eqnarray}

But then, \begin{eqnarray} \displaystyle{\sum\limits_{k=N+1}^\infty \frac{a_k}{1+\epsilon} < \sum\limits_{k=N+1}^\infty \frac{a_n}{1+a_n}} \end{eqnarray}

Since $\sum a_n$ diverges, so does $\displaystyle{\sum\limits_{k=N+1}^\infty \frac{a_k}{1+\epsilon}}$.

By the comparison test, $\displaystyle{\sum\limits_{k=N+1}^\infty \frac{a_n}{1+a_n}}$ also diverges.

• See this for a more general problem (you need not assume $(a_n)$ has limit $0$). – David Mitra Jun 5 '14 at 12:04
• Your proof looks ok; but you don't want to sum all the way to $\infty$. Instead, say $\sum_{k=N+1}^{N+m}{a_k\over 1+\epsilon}\le\sum_{k=N+1}^{N+m}{a_k\over 1+a_k}$ for all $m\ge1$. Then argue that your sum diverges. Note that if you appeal to the Comparison Test, all this is unnecessary; all you need is your initial inequality. – David Mitra Jun 5 '14 at 12:22

## 2 Answers

If you use Limit comparison test then you get $$\lim\limits_{n\to\infty}\frac{\frac{a_n}{1+a_n}}{a_n}=\lim\limits_{n\to\infty}\frac{1}{1+a_n}=1.$$ This means that the series $\sum a_n$ and $\sum\frac{a_n}{1+a_n}$ have same character. So $\sum\frac{a_n}{1+a_n}$ diverges.

Your proof is right!

In fact, $\lim a_n = 0$ can be removed.

If $a_n\geq1$, then

$$\frac{a_n}{1+a_n}\geq\frac12$$

so, if there are infinite $n$ such that $a_n\geq1$, then $\displaystyle{\sum \frac{a_n}{1+a_n}}$ diverges.

If there exists $N$ such that $\forall n\geq N$, $a_n\leq1$, then

$$\frac{a_n}{1+a_n}\geq\frac{a_n}2$$

we have

$$\sum_{n=N}^\infty\frac{a_n}{1+a_n}\geq \sum_{n=N}^\infty\frac{a_n}2$$

• @Alex $2>1$; $2/(1+2)=2/3>1/2$. – Julián Aguirre Jun 5 '14 at 15:48