# Convergence of series with alternating terms

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.

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:

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

• Is there a way to solve the inequality without using calculus ? – Al-C Jan 11 at 12:56
• There might be but I don't see an easy way out - the calculus approach from above is probably the easiest/fastest. – StackTD Jan 11 at 13:11

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.