Positive series problem: $\sum\limits_{n\geq1}a_n=+\infty$ implies $\sum_{n\geq1}\frac{a_n}{1+a_n}=+\infty$

Question

Let $\sum\limits_{n\geq1}a_n$ be a positive series, and $\sum\limits_{n\geq1}a_n=+\infty$, prove that: $$\sum_{n\geq1}\frac{a_n}{1+a_n}=+\infty.$$

Answer 1

Proving the contrapositive statement seems cleaner to me. Suppose $\sum{a_n\over 1+{a_n}}$ converges. Then ${a_n\over 1+{a_n}}\rightarrow 0$. This implies that ${a_n}$ is eventually less than one, so ${a_n\over2}\le {a_n\over a_n+1}$ for $n$ sufficiently large. The comparision test then shows that $\sum a_n$ converges.

Answer 2

I would argue by cases.

Case 1. $a_n\ge1$ for infinitely many $n\in\mathbb N$.

In this case, for each such $n$ we have $\frac{a_n}{1+a_n}=1-\frac{1}{1+a_n}\ge1-\frac12$, from which the claim easily follows.

Case 2. $a_n\ge1$ for only finitely many $n\in\mathbb N$.

In this case for every other $n$ we have $a_n&lt1$ and thus $\frac{a_n}{1+a_n}\ge\frac{a_n}{1+1}=\frac{a_n}2$. Since finitely many terms can't affect the convergence/divergence of a series, this will also diverge. (Since $\sum\limits_{n=1}^\infty\frac{a_n}{2}$ does.)

Answer 3

Alternatively, split the problem up in cases:

1, If there is a natural $N\in\mathbb{N}$ s.t $a_n\leq{1}$ for $n\geq N$, what can you conclude?

2, If (1) is not true, for every natural $N$ we can find a $n\geq{N}$ s.t $a_n>1.$ Now passing to a subsequence and comparing with a series with each term equal to a constant (more precisely $1/2$ or lower), the result follows.

Answer 4

You can suppose that $(a_n) \rightarrow 0$ (if not the problem is trivial). Then what can you say asymptotically about $\frac{a_n}{1+a_n}$ ?

Answer 5

We can divide into cases:

1. If a(n) has limit zero : It is lower than 1 for all n bigger than n0, then we can compare with a(n)/2 which is lower than a(n)/(1+a(n)).

2. If a(n) has limit different to zero , also a(n)/1+a(n) and then the series diverges

3. If a(n) is not bounded it ha a subsequence that converges to infinite, then a(n)/1+a(n) converges to 1 then the series diverges to infinite.

4. If a(n) is bounded , we can take a subsequence that is convergent.

If it does not converges to zero also the sequence a(n)/1+a(n). If all subsequences converge to zero ,then also a(n) and we can apply 1.

Answer 6

Suppose $\sum{a_n\over 1+{a_n}}$ converges. Then $\frac{a_n}{1+a_n}\to 0$. It is easy to see that: $$\lim_{n\to\infty}a_n=0\iff \lim_{n\to\infty}\frac{a_n}{1+a_n}=0.$$ (let $b_n=\frac{a_n}{1+a_n}$, then $a_n=\frac{b_n}{1-b_n}$ )

So we have $$\lim_{n\to\infty}\frac{\frac{a_n}{1+a_n}}{a_n}=1,$$ by comparision test of positive series, the series $\sum a_n$ converges, it is a contradiction.

Answer 7

For $a,b>0$ observe that $$ \frac{a}{a+1}+\frac{b}{b+1} > \frac{a+b}{a+b+1},\qquad (\star) $$ since $$ \frac{a}{a+1}+\frac{b}{b+1} - \frac{a+b}{a+b+1}=\bigl((a+1)+(b+1)\bigr)\left(\frac{1}{a+b+1}-\frac{1}{ab+a+b+1}\right). $$ By iteratively applying $(\star)$ to we find that the inequality extends to sequences as well, yielding $$ \frac{a_1}{a_1+1}+\frac{a_2}{a_2+1}+\cdots+\frac{a_n}{a_n+1}>\frac{a_1+\cdots+a_n}{1+a_1+\cdots+a_n}.\qquad (\star\star) $$ Now consider a divergent series $\sum a_n$ of positive reals. Since the series diverges, there exists an infinite sequence of indices such that $$ a_{i_k}+\cdots+a_{j_k}\geq 1,\qquad i_1<j_1<i_2<j_2<\cdots. $$ Applying $(\star\star)$ to each subsequence $a_{i_k},\ldots,a_{j_k}$ in turn yields $$ \sum_{n=i_k}^{j_k}\frac{a_n}{1+a_n}>\frac{1}{2}. $$ Thus the series $ \sum\frac{a_n}{1+a_n} $ diverges, since it can be written as a sum over infinitely many subsequences each of which contributes at least $\tfrac12$ to the sum.