Examine the uniform convergence of the series $$\sum^{\infty}_{n=1}\frac{1}{\sqrt{x+n}}$$ if $x \in [0, \infty)$

Which series should I choose in Weierstrass M-test to show that is divergent?

Of course $$\sum^{\infty}_{n=1}\frac{1}{\sqrt{n}} = \sum^{\infty}_{n=1}\frac{1}{n^{\frac{1}{2}}}$$ and this series diverges (and for $x = 0$ function series is divergent), yet $$\frac{1}{n^{\frac{1}{2}}} \geqslant \frac{1}{\sqrt{x+n}}$$ for positive x, so it do not work.

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    $\begingroup$ The series is nowhere convergent. $\endgroup$ – Mark Viola Aug 23 '15 at 22:08
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    $\begingroup$ $$\sum^{\infty}_{n=1}\frac{1}{\sqrt{x+n}}\geqslant\sum^{\infty}_{n=1}\int_{x+n}^{x+n+1}\frac{1}{\sqrt{t}}dt=\int_{x+1}^\infty\frac{1}{\sqrt{t}}dt=+\infty$$ $\endgroup$ – Did Aug 23 '15 at 22:46

Hint: For $n > x$ you can approximate $\sqrt{x + n} \le \sqrt{2}\sqrt{n}$


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