Given the series: $$\sum_{n=1}^\infty(-1)^{n-1} \frac{\sqrt{n}}{n+4}, $$

how to determine if it's divergent or convergent?

The problem seems harder than it looks. It's not absolutely convergent, as can be easily shown by limit comparison test. I'm thinking of using Alternating Series Test, but I'm not sure if the absolute value of the expression is decreasing.


closed as off-topic by Simply Beautiful Art, Did, Namaste, José Carlos Santos, Xander Henderson Feb 26 '18 at 1:46

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Yes it is an alternating series and $a_n\to 0$ is monotonically decreasing indeed

$$f(x)=\frac{\sqrt{x}}{x+4}\implies f'(x)=\frac{4-x}{2\sqrt x(x+4)^2}<0$$

for $x>4$.


$$\frac{\sqrt n}{n+4}\ge\frac{\sqrt{n+1}}{n+5}\iff(n+5)^2n\ge(n+4)^2(n+1)\iff$$

$$n^3+10n^2+25n\ge n^3+9n^2+24n+16\iff n^3+n^2+n\ge16$$

and the above happens for$\;n\ge 3\implies\;$ the sequence of the series (without $\;(-1)^n\;$ is monotonic descending to zero and thus yours is a Leibniz series which, of course, converges conditionally.


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