# Absolutely Convergent, Conditionally Convergent or Divergent?

$$\sum_{k=0}^{\infty} (-1)^{k+1}\frac{\sqrt{k}}{k+1}$$

This problem is asking me to prove if this series is absolutely convergent, conditionally convergent or divergent, but I don't know how to start when I begin with the absolute convergence. Can anyone give me a hint on how to do it? Thanks

• no, is the square root of k, not the whole thing – user354632 Jul 26 '16 at 11:57
• So my edit in the post is correct? Anyway, you should visit the Help center for formatting tips math.stackexchange.com/help/notation – Yuriy S Jul 26 '16 at 11:58

For conditionally convergence use the Leibniz convergence theorem, the sequence is decreasing and converges to zero. But the series does not converge absolute:$$\frac{\sqrt{k}}{k+1}\geq \frac{1}{k+1}$$ which does diverge.
Hint. One may write, as $k \to \infty$, using a Taylor series expansion, \begin{align} \frac{\sqrt{k}}{k+1}&=\frac1{\sqrt{k}}\frac1{1+\frac1{\sqrt{k}}} \\&=\frac1{\sqrt{k}}\left(1-\frac1{\sqrt{k}}+O\left(\frac1k\right)\right) \\&=\frac1{\sqrt{k}}-\frac1k+O\left(\frac1{k^{3/2}}\right) \end{align} thus the given series $$\sum_{k\ge k_0}\frac{(-1)^k\sqrt{k}}{k+1}=\sum_{k\ge k_0}\frac{(-1)^k}{\sqrt{k}}-\sum_{k\ge k_0}\frac{(-1)^k}k+\sum_{k\ge k_0}O\left(\frac1{k^{3/2}}\right)$$ is conditionally convergent, not absolutely convergent.
• @user354632 As given in my answer above, by comparison, as $k \to \infty$, $\frac{\sqrt{k}}{k+1} \sim \frac1{\sqrt{k}}$ and the series is not absolutely convergent. – Olivier Oloa Jul 26 '16 at 12:12