# Absolute and Uniform Convergence of a Series

$$\sum_{k=1}^\infty \frac{(-1)^k}{\sqrt{k-1}+(-1)^k}$$ I put it on wolfram alpha and it shows that converges, but I think the analysis has to be different, because I need to know if the series converge absolutely or uniform or conditional and wolfram doesn't show it. Any help you can give will be greatly appreciated

• We cannot speak about uniform convergence here. – hamam_Abdallah May 31 '17 at 1:57

The series fails to converge even conditionally since

$$\frac{(-1)^k}{\sqrt{k-1} + (-1)^k} = \frac{(-1)^k(\sqrt{k-1} - (-1)^k)}{(\sqrt{k-1} + (-1)^k)(\sqrt{k-1} - (-1)^k)} = \frac{(-1)^k \sqrt{k-1}}{k-2} - \frac{1}{k-2}$$

and we have divergence of

$$\sum_{k=3}^\infty \frac{1}{k-2}$$

and convergence by the Dirichlet test of

$$\sum_{k=3}^\infty \frac{(-1)^k\sqrt{k-1}}{k-2}$$

• This presents the quintessential example of a divergent alternating series for which the absolute value of general terms approach zero, thereby showing the importance of the monotonicity condition in Leibniz's test. (+1) – Mark Viola May 31 '17 at 2:44
• @MarkViola: Glad somebody reads this stuff. Congratulations BTW – RRL May 31 '17 at 2:48
• Ah, thank you! I have been told all of the secrets of the universe ... and I get a t-shirt! What's better than that? – Mark Viola May 31 '17 at 2:51
• Sorry but I don't understand how to get this step $$\frac{(-1)^k(\sqrt{k-1} - (-1)^k)}{(\sqrt{k-1} + (-1)^k)(\sqrt{k-1} - (-1)^k)} = \frac{(-1)^k \sqrt{k-1}}{k-2} - \frac{1}{k-2}$$ can you explain it please? @RRL – C. Ballez Sep 21 '17 at 19:53
• Sure. We know $(a-b)(a+b) = a^2 - b^2$. Applying this to the denominator we get $(\sqrt{k-1} + (-1)^k)(\sqrt{k-1} - (-1)^k) = (\sqrt{k-1})^2 - (-1)^{2k} = k-1 - ((-1)^2)^k = k - 1 -1^k = k-2$.// In the numerator we get $(-1)^k(\sqrt{k-1} - (-1)^k) = (-1)^k\sqrt{k-1} - (-1)^{2k} = (-1)^k\sqrt{k-1} - 1$. This rationalization is applicable to all terms with index $k \geqslant 3$ where the denominator is not $0$ and it is enough to consider the sum starting with $k=3$ to examine convergence / divergence. – RRL Sep 21 '17 at 20:24

hint for absolute

$$|u_k|\ge \frac {1}{\sqrt {k-1}+1}$$

$$\frac {1}{\sqrt {k-1}+1}\sim \frac {1}{\sqrt {k}}$$

$\sum \frac {1}{\sqrt {k}}$ is divergent thus

$\sum u_k$ is not absolutely conv.