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I tried all the theorems, that I knew in analysis, to know if the mentioned series converge but none of them is relevant except one: The Ratio Test for Series, but unfortunately this is not working as the limit is neither less nor greater than $1$, i.e., $$\lim_{n \to \infty} \dfrac{|a_{n+1}|}{|a_n|} = \dfrac{\Big|\ln\Big(1+\dfrac{x}{n+1}\Big) - \dfrac{x}{n+1} \Big|}{\Big|\ln\Big(1+\dfrac{x}{n}\Big) - \dfrac{x}{n} \Big|} = 1.$$

How can I prove if the series $$\sum_{k=1}^{\infty} \left[\ln\left(1+\frac{x}{k}\right) - \frac{x}{k} \right]$$ is convergent or not?

Thank you.

Edit - $x \in \mathbb{R} - {\{0, -1, -2, \dots}\}$. Same domain for $\Gamma (x)$. Please read this answer.

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    $$-\frac{x^2}{2k^2}\leqslant\ln\left(1+\frac{x}{k}\right) - \frac{x}{k} \leqslant0\qquad(k\geqslant x)$$ – Did Apr 16 '16 at 23:14
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    IOW Taylor's theorem on $\ln(1+u)$. – anon Apr 16 '16 at 23:15
  • @Did - I got the question (OP) when I read this. No obligation of $k \ge x$ for $x$ exists except for $x$ is a nonzero nonnegative real number. Will it change the convergancy? –  Apr 16 '16 at 23:25
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    @Edi For every fixed $x$, $k \geq x$ holds eventually (that is, for sufficiently large $k$) so it's fine. – MT_ Apr 16 '16 at 23:30
  • @Soke, Yes! Thanks a lot. Problem solved! :)) –  Apr 16 '16 at 23:32
  • "Edit - $x \in \mathbb{R} - {{0, -1, -2, \dots}}$" No, rather $x \in \mathbb{R}$, $x>-1$. – Did Apr 17 '16 at 07:08

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