Let $\sum\limits_{n\geq1}a_n$ be a positive series, and $\sum\limits_{n\geq1}a_n=+\infty$, prove that: $\sum_{n\geq1}\frac{a_n}{1+a_n}=+\infty.$
Positive series problem: $\sum\limits_{n\geq1}a_n=+\infty$ implies $\sum_{n\geq1}\frac{a_n}{1+a_n}=+\infty$
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0fantastic. thanks for referring to the book. – 2013-05-18
5 Answers
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.
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0@Fundamental Go ahead. I won't complain. – 2015-01-27
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<1 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.)
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.
You can suppose that $(a_n) \rightarrow 0$ (if not the problem is trivial). Then what can you say asymptotically about $\frac{a_n}{1+a_n}$ ?
We can divide into cases:
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)).
If a(n) has limit different to zero , also a(n)/1+a(n) and then the series diverges
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.
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.
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0We can restrict to the last two cases but I started with the most common to think about. – 2012-04-14