We need to determine the convergence or divergence of the infinite series $\sum_{k=1}^{\infty} \frac{1}{k\sqrt{k}}$. If it converges, find its value.

1. Problem Description

We need to determine the convergence or divergence of the infinite series $\sum_{k=1}^{\infty} \frac{1}{k\sqrt{k}}$. If it converges, find its value.

2. Solution Steps

The given series is $\sum_{k=1}^{\infty} \frac{1}{k\sqrt{k}}$.

We can rewrite the term inside the summation as $\frac{1}{k\sqrt{k}} = \frac{1}{k \cdot k^{1/2}} = \frac{1}{k^{3/2}}$.

Thus, the series can be written as $\sum_{k=1}^{\infty} \frac{1}{k^{3/2}}$. This is a $p$-series with $p = \frac{3}{2}$.

A $p$-series of the form $\sum_{n=1}^{\infty} \frac{1}{n^p}$ converges if $p > 1$ and diverges if $p \le 1$.

In this case, $p = \frac{3}{2} > 1$, so the series converges.

To find the sum, we use the definition of the Riemann zeta function:

$\zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s}$ for $s > 1$.

In our case, $s = \frac{3}{2}$.

So the sum is $\zeta(\frac{3}{2})$. The Riemann zeta function $\zeta(\frac{3}{2})$ has no known closed-form expression. Its approximate value is $\zeta(\frac{3}{2}) \approx 2.612$.

3. Final Answer

$\zeta(3/2)$