# If $\sum_{n = 1}^\infty {{a_n}}$ converges, then is $\sum_{n = 1}^\infty (1+a_n)^{-1}$ a convergent series?

If $\sum\limits_{n = 1}^\infty {{a_n}}$ is convergent (with ${a_n} > 0$, $\forall n\in\mathbb{Z}$), then is $\sum\limits_{n = 1}^\infty {\left( {\frac{1}{{1 + {a_n}}}} \right)}$ is a convergent series?

I tried with the quotient criterion and comparison theorem no avail. But I figured something was this: $${\frac{1}{{1 + {a_n}}}}<1$$ but still I find it useful. Any suggestions please.

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What's $\lim a_n$, and what's $\lim \frac{1}{1+a_n}$? –  Daniel Fischer Dec 3 '13 at 21:50
Good! Thanks Daniel –  mathsalomon Dec 3 '13 at 21:55
For example ( a 'counter...' one) $\displaystyle a_{n} \equiv {1 \over n^{2}}$. –  Felix Marin Dec 4 '13 at 4:16

If $\sum a_n$ converges, then $a_n \to 0$, therefore $\frac{1}{1+a_n} \to 1$ and $\sum \frac{1}{1+a_n}$ diverges.

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If $\sum a_n <\infty$ then of course $\lim a_n =0$. Therefore $\lim \frac{1}{1+a_n} =1$, so second series does not satisfy the term-test http://en.wikipedia.org/wiki/N-th_term_test

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@nik. The fastest in the west haha.. Many thanks nik –  mathsalomon Dec 3 '13 at 21:54

If the first series converges, then surely $a_n \rightarrow 0$. If that's true, can the terms in the second series also approach zero? If the second series is to converge, then they must. Said differently, if they don't, then the second series can't converge.

So if $\lim_{n \rightarrow \infty} a_n = 0$, what is $\lim_{n \rightarrow \infty} \frac{1}{1+a_n}$ ?

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the limit is one. thanks –  mathsalomon Dec 3 '13 at 21:59

A more interesting situation is to consider $$\tag 1 \sum_{i\geqslant 1}\frac{a_i}{1+a_i}$$

If $a_i>0$, and $$\tag 2 \sum_{i\geqslant 1} a_i$$ exists, we have that $\dfrac{1}{1+a_i}\to 1$. Since this is the quotient of the $i$-th term of $(1)$ with that of $(2)$, this means that $(1)$ converges if $(2)$ does. But if $(1)$ converges, from $\dfrac{a_i}{1+a_i}\to 0$ and $1+a_i>0$ we get that $a_i\to 0$. This gives again a quotient of $1$ and by comparison $(2)$ converges.

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