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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

juantheron
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  • Presumably, we are treating $S$ as a function on the positive integers, and asking for its range. Its range is the set $${,1,1+(1/\sqrt2),1+(1/\sqrt2)+(1/\sqrt3),\dots,}$$ The lower bound is 1, and there is no upper bound, since the series diverges. – Gerry Myerson Sep 29 '13 at 05:45

1 Answers1

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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 $S$.

Falang
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