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I am trying to prove whether the following alternating series is convergent or not. $$\sum_{n=1}^\infty (-1)^{n}\frac{\log{n}}{\sqrt{n}}$$ One condition that must be satisfied is that $$ \|a_{n+1}\|\le\|a_{n}\|$$ This leads to the following inequality $$\log(n+1)^\frac{1}{\sqrt{n+1}}\le\log(n)^\frac{1}{\sqrt{n}}$$ Then $$\log\left[(n+1)^\frac{1}{\sqrt{n+1}}/(n)^\frac{1}{\sqrt{n}}\right]\le0$$ And if I am correct $$\left[(n+1)^\frac{1}{\sqrt{n+1}}/(n)^\frac{1}{\sqrt{n}}\right]\le1$$

I have plotted the inequality and checked that it is true when $n>7$. However I cannot prove it.

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Hint

$$\frac{\mbox{d}}{\mbox{d}x}\left(\frac{\log x}{\sqrt{x}}\right) = \frac{2-\log x}{2x^{3/2}} < 0 \quad\mbox{ for }\quad x \ldots$$


$$\|a_{n+1}\|\le\|a_{n}\|$$

Side remark; I guess you want absolute values rather than norms:

$$\left|a_{n+1}\right|\le\left|a_{n}\right|$$

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  • $\begingroup$ Is there a way to solve the inequality without using calculus ? $\endgroup$ – Al-C Jan 11 at 12:56
  • $\begingroup$ There might be but I don't see an easy way out - the calculus approach from above is probably the easiest/fastest. $\endgroup$ – StackTD Jan 11 at 13:11
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Consider $x^{-1/2}\log\, x$. Its derivative is $-\frac 1 2 x^{-3/2}\log\, x+x^{-3/2}$. Since $\log\, x >2$ for $x$ sufficiently large it follow that the function is decreasing in some interval of the type $(a,\infty$). If you ignore the first few terms of the given series the terms are decreasing in absolute value and we can apply alternating series test.

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