Closed form for : $L=1+{1 \over 2^{1/2}}+{1 \over 3^{1/2}}+{1 \over 4^{1/2}}+\cdots$ exist?

Question

$$S=1+{1 \over 2^2}+{1 \over 3^2}+{1 \over 4^2}+\cdots\tag1$$

it is known $$S={\pi^2\over 6}$$

How about the series

$$L=1+{1 \over 2^{1/2}}+{1 \over 3^{1/2}}+{1 \over 4^{1/2}}+\cdots\tag2$$

Does L converges?. If it does. Has its got any closed form? I look around but could find anywhere.

Suppose that it is found online. Can anyone show how its closed form is derived.

Answer 1

No, the series does not converge. It has the same convergence as integral: $$\int_1^{\infty} \frac{1}{\sqrt{x}}dx = \lim_{x \to \infty} \frac{\sqrt{x}}{2} - \frac{1}{2} = \infty$$ By integral test your sum and my integrals have same convergence.

Answer 2

$$L > 1 + \frac 12 + \frac 13 + \frac 14 + \cdots \to \infty$$

Answer 3

Consider the integral $\displaystyle\int_{1}^{\infty}\dfrac{1}{x^{1/2}}dx$, it is divergent, and hence by Integral Test, the series is divergent as well.

Answer 4

Since $\frac{1}{\sqrt{n}} \ge \frac{1}{n}$ for all $n$, the series

$ \sum_{n=1}^{\infty}\frac{1}{\sqrt{n}}$ diverges, by the comparison test.