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2

$$F(s) = a_1+\sum_{n=2}^\infty a_n n^{-s}$$ if $a_1 \ne 0$ : we can consider the Dirichlet series $F(s)/a_1$ so without loss of generality we suppose $a_1 = 1$. $F(s)$ is supposed to be convergent at $s=\sigma+it$ so $a_n = o(n^{\sigma})$ so $F(s)$ is absolutely convergent for $Re(s) > \sigma+1 = \sigma_0$ and when $Re(s) \to \infty$ : $F(s) = ... 3 It is possible if$1/F$has a representation as a Dirichlet series. To obtain the coefficients, it is useful to know how the product of two Dirichlet series looks. Let hence $$F(s) = \sum_{n = 1}^{\infty} \frac{a(n)}{n^s}\quad \text{and}\quad G(s) = \sum_{n = 1}^{\infty} \frac{b(n)}{n^s}$$ where both series are absolutely convergent for$\operatorname{Re} ...

1

As $\sum x_n$ converges absolutely , $$x_n \to 0 \implies 1 +x_n \to 1 \implies |1+x_n|\to 1 \implies |1+x_n| > \frac 1 2$$ for all $n >$ some $k$ Hence $\sum_{n>k}|\frac {x_n}{1+x_n}|\leq 2\sum _{n>k}|x_n|<\infty$

2

We have $x_n \to 0$. Hence there is some $N$ such that $n \geq N$ implies $x_n > -1/2$. Therefore when $n \geq N$, we have $$\left| \frac{x_n}{1 + x_n}\right| = \frac{|x_n|}{1 + x_n} \leq 2|x_n|.$$ Thus the series $\sum \frac{x_n}{1 + x_n}$ converges absolutely by comparison with $\sum |x_n|$.

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$\sum|x_n|$ converges $\implies x_n\to 0\implies -1/2\le x_n$ for $n$ large enough $\implies 1/2\le 1+x_n$ for $n$ large enough $\implies |x_n/(1+x_n)|\le 2|x_n|$ for $n$ large enough.

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If you want to show that $\sum \frac{x_n}{1+x_n}$ converges, define $f(x)=\frac{x}{1+x}$, it is a continuous function. Since $x_n$ converges towards a number different of $-1$, so is $f(x_n)$.

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