Does the following series converge? Does the following series converge ?
$$1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{n}}+\ldots$$
Let, $u_{n}=\frac{1}{\sqrt{n}}$
$\lim\limits_{n\to\infty}{u_{n}}=\lim\limits_{n\to\infty}\frac{1}{\sqrt{n}}$
$\lim\limits_{n\to\infty}{u_{n}}= 0$
 A: $$\sum_{n = 1}^\infty \frac{1}{\sqrt{n}} = \sum_{n = 1}^\infty \frac 1{n^{(1/2)}}$$
By the $p-\text{series test},\;$ if the exponent $p$ of $n$ in the denominator is $p \leq 1$, the series diverges. Here, that exponent $p = 1/2 \lt 1$, hence, the series diverges. 
A: $1+\dfrac{1}{2}+\dfrac{1}{3}+\cdot\cdot\cdot+\dfrac{1}{n}$ is diverges when $n\to\infty$, $\dfrac{1}{\sqrt{n}}>\dfrac{1}{n}$, so the series is diverges when $n\to\infty$.
A: Hint: Use the integral test, 
$$ \int_{1}^{\infty}\frac{1}{\sqrt{x}} dx. $$ 
Added: Here is the main result. 

Theorem: Suppose $f$ is continuous, positive, decreasing function on $[1,\infty)$ and let $a_n=f(n)$. Then 
$(a)$ if $\int_{1}^{\infty}f(x) dx$ is convergent, then $\sum_{n=1}^{\infty} a_n $ is convergent.  
$(b)$ if $\int_{1}^{\infty}f(x) dx$ is divergent, then $\sum_{n=1}^{\infty} a_n $ is divergent.

Note: Now, you can see, for any series of the form $\sum_{n=1}^{\infty}\frac{1}{n^p}$, we can find the condition on $p$ for which the series converges or diverges by considering the integral
$$ \int_{1}^{\infty}\frac{1}{x^p} dx. $$  
A: This series diverges from the $p$-test. If you're unfamiliar with this approach to the problem, check this link: http://www.sosmath.com/calculus/improper/testconv/testconv.html. This is a more elementary method. Rather than using the integral test like Sean Gomes suggested.
A: No.
In general, the $n$-th term of a sequence tending to zero just not imply that the corresponding infinite series converges.
One way of proving divergence in this example is by using the integral test.
A: General purpose tests are really useful for such problems (and there are answers with those).
Here is a different way of proving divergence.
$$\sqrt{k+1} - \sqrt{k} = \frac{1}{\sqrt{k+1} + \sqrt{k}} \le \frac{1}{2\sqrt{k}}$$
Thus
$$\sqrt{n+1} - 1 = \sum_{k=1}^{n} (\sqrt{k+1} - \sqrt{k}) \le \sum_{k=1}^{n} \frac{1}{2\sqrt{n}}$$
and so
$$\sum_{k=1}^{n} \frac{1}{\sqrt{n}} \ge 2(\sqrt{n+1} - 1)$$
