Show that the sequence $1+1/\sqrt{2}+1/\sqrt{3}+...+1/\sqrt{n}$ is unbounded I need to show that $1+1/\sqrt{2}+1/\sqrt{3}+...+1/\sqrt{n}$ is unbounded. So I can show that for $S_n=1+1/2=1/3+....+1/n$, $a_n > S_n$, and $S_n$ is not bounded above, but how do I show that $a_n$ is not bounded below?
 A: Here's another proof:
$\frac{1}{n}\sum_{i=0}^n\frac{1}{\sqrt{i}}$ is the average value of the set $\{1,\ldots,\frac{1}{\sqrt{n}}\}$. The average value of a set is always bigger than its smallest element, so we get
$$\frac{1}{n}\sum_{i=0}^n\frac{1}{\sqrt{i}}\geq\frac{1}{\sqrt{n}}$$
Multiplying both sides by $n$ tells us that $\sum_{i=0}^n\frac{1}{\sqrt{i}}\geq\sqrt{n}$ which diverges.
A: Hint: $$\left(\sum_{i=1}^n \frac{1}{\sqrt{i}}\right)^2> \sum_{i=1}^n \frac{1}{i}$$ (If you do not wish to use integral test, that is.)
A: Hint: Compare the sequence with
$$S_n = \sum_{k=1}^n \frac{1}{k} $$
A: $\sum_{k=1}^n \frac1{\sqrt{k}}
\ge \sum_{k=1}^n \frac1{\sqrt{n}}
=\frac{n}{\sqrt{n}}
=\sqrt{n}
$.
More generally,
if $a < 1$,
$\sum_{k=1}^n \frac1{k^a}
\ge \sum_{k=1}^n \frac1{n^a}
=\frac{n}{n^a}
=n^{1-a}
\to \infty
$.
A: Assume that $$1+\dfrac1{\sqrt{2}}+\dfrac1{\sqrt{3}}+…+\dfrac1{\sqrt{n}}\gt\sqrt{n}$$ Then $$1+\dfrac1{\sqrt{2}}+\dfrac1{\sqrt{3}}+…+\dfrac1{\sqrt{n}}+\dfrac1{\sqrt{n+1}}\gt\sqrt{n}+\dfrac1{\sqrt{n+1}}\gt\sqrt{n+1}.$$ Hence by Mathematical Induction.....
A: Here is an approach. Observe that 
$$
\frac1{\sqrt{x}}\leq \frac1{\sqrt{k}},\qquad x\in [k,k+1],\quad k=1,2,3,\cdots.
$$
Integrating both sides from $x=k$ to $x=k+1$ gives
$$
\int_k^{k+1}\frac1{\sqrt{x}}\:dx\leq \frac1{\sqrt{k}}=\int_k^{k+1}\frac1{\sqrt{k}}\:dx
$$ then summing gives
$$
\int_1^{n}\frac1{\sqrt{x}}\:dx\leq 1+\dfrac1{\sqrt{2}}+\dfrac1{\sqrt{3}}+…+\dfrac1{\sqrt{n}}
$$ evaluating the left hand side 
$$
2\sqrt{n}-2\leq 1+\dfrac1{\sqrt{2}}+\dfrac1{\sqrt{3}}+…+\dfrac1{\sqrt{n}}
$$ and letting $n \to \infty$ leads to the announced result.
A: $S_2=1+\frac{1}{\sqrt{2}} >\frac{2}{\sqrt{2}}$
$S_3 = 1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}} > \frac{3}{\sqrt{3}}$
$\vdots$
$S_N > \frac{N}{\sqrt{N}} = \sqrt{N} \to \infty$ as $N \to \infty$ 
A: A really twisted form using Cesaro-Stolz:
$$
\lim_{n\to\infty}\frac{1+1/\sqrt{2}+1/\sqrt{3}+...+1/\sqrt{n}}{\sqrt n}=
\lim_{n\to\infty}\frac{1/\sqrt{n+1}}{\sqrt{n+1}-\sqrt n} =
\lim_{n\to\infty}\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}} = 2.
$$
