Range of $S = \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+..............+\frac{1}{\sqrt{n}}$ If $\displaystyle S = \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+..............+\frac{1}{\sqrt{n}}$ and $n\in \mathbb{N}$. Then Range of $S$ is
$\underline{\bf{My \;\; Try}}$:: For Lower Bond:: $\sqrt{n}\geq \sqrt{r}\;\; \forall r\in \mathbb{N}$
So $\displaystyle \frac{1}{\sqrt{r}}\geq \frac{1}{\sqrt{n}}$ Now Adding from $r = 1$ to $r=n$
$\displaystyle \sum_{r=1}^{n}\frac{1}{\sqrt{r}}\geq \frac{n}{\sqrt{n}} = \sqrt{n}\Rightarrow \sum_{r=1}^{n}\frac{1}{\sqrt{r}}\geq\sqrt{n}$
Now How Can i found Upper Bond , Help Required.
Thanks

When we say a series is bounded, we mean it is bounded by a CONSTANT. As Gerry wrote above,  lower bound is $1$ and there is no upper bound.
If you are familiar with the Harmonic series, you can see each term in $S$ greater equal than each term of the Harmonic series. We all know the Harmonic series diverges and hence there is no upper bound for $S$.
