Prove by induction that $\sum_{i=1}^n \frac{1}{\sqrt i} > \sqrt n$ for all integers $n \ge 2$ Having trouble with the concept of proving an inequality.
Q: Prove by induction that $\sum_{i=1}^n \frac{1}{\sqrt i} > \sqrt n$ for all integers $n \ge 2$
Here is what I have so far:
Basis ($n=2$)
$\sum_{i=1}^n \frac{1}{\sqrt i} = \frac{1 + \sqrt 2}{\sqrt 2} > \frac{1 + 1}{\sqrt 2} = \sqrt 2$
Inductive Step
(IH): Let $k \ge 2$ be an integer and suppose that $\sum_{i=1}^k \frac{1}{\sqrt i} > \sqrt k$
We want to prove that $\sum_{i=1}^{k+1} \frac{1}{\sqrt i} > \sqrt {k+1}$
So,
$$\sum_{i=1}^{k+1} \frac{1}{\sqrt i} = \left( \sum_{i=1}^k \frac{1}{\sqrt i} \right) + \frac {1}{\sqrt {k+1}}$$
$$\sum_{i=1}^{k+1} \frac{1}{\sqrt i} > \sqrt k + \frac {1}{\sqrt {k+1}}$$
$$\sum_{i=1}^{k+1} \frac{1}{\sqrt i} > \frac{\sqrt{k+1} \sqrt k}{\sqrt{k+1}} + \frac{1}{\sqrt{k+1}}$$
$$\sum_{i=1}^{k+1} \frac{1}{\sqrt i} > \frac{\sqrt{k+1} \sqrt k + 1}{\sqrt{k+1}}$$
And now I do not understand where to go from here to get my proof. I don't know what I'm supposed to compare here?
 A: Induction is not at all necessary here: if $1\le  i\le n$, so $1\le \sqrt i \le \sqrt n$, whence $\dfrac 1{\sqrt n}\le \dfrac 1{\sqrt i}\le 1$, and
$$ \sum_{i=1}^n \frac{1}{\sqrt n}=\frac n{\sqrt n}=\sqrt n\le \sum_{i=1}^n \frac{1}{\sqrt i} \le \sum_{i=1}^n 1=n. $$
Furthermore, if $n\ge 2$, all but one of the inequalities $\frac1{\sqrt i}<1$  and all but one of  $\frac1{\sqrt n}<\frac1{\sqrt i}$ are strict, so  really 
$$\sqrt n < \sum_{i=1}^n \frac{1}{\sqrt i} <n. $$
A: Based on your derivation, it suffices to show that 
$$
\sqrt{k+1}\sqrt{k} + 1 \ge k+1.
$$
This can be shown as follow:
$$
(k+1)k = k^2 + k > k^2.
$$
Then you're done!
A: To make the proof work,
you need to show that
$\sqrt{n}+\dfrac1{\sqrt{n+1}}
\gt \sqrt{n+1}
$.
This is the same as
$\begin{array}\\
\dfrac1{\sqrt{n+1}}
&\gt \sqrt{n+1}-\sqrt{n}\\
&=(\sqrt{n+1}-\sqrt{n})\dfrac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}\\
&=\dfrac{1}{\sqrt{n+1}+\sqrt{n}}\\
\end{array}
$
which is obvious.
You can get a more precise result
by noting that
$\dfrac1{\sqrt{n+1}}
=\dfrac{1}{\sqrt{n+1}+\sqrt{n}}
$
implies that
$\dfrac{1}{2\sqrt{n+1}}
\lt \sqrt{n+1}-\sqrt{n}
\lt \dfrac{1}{2\sqrt{n}}
$.
Summing the left side,
$\sum_{n=1}^m \dfrac{1}{2\sqrt{n+1}}
\lt \sum_{n=1}^m (\sqrt{n+1}-\sqrt{n})
=\sqrt{m+1}-1
\lt \sqrt{m+1}
$
so that
$\sum_{n=2}^{m+1} \dfrac{1}{2\sqrt{n}}
\lt \sqrt{m+1}
$
or
$\sum_{n=2}^{m} \dfrac{1}{\sqrt{n}}
\lt 2\sqrt{m}
$
so
$\sum_{n=1}^{m} \dfrac{1}{\sqrt{n}}
\lt 2\sqrt{m}+1
$.
Summing the right side,
$\sum_{n=1}^{m-1} \dfrac{1}{2\sqrt{n}}
\gt \sum_{n=1}^{m-1} (\sqrt{n+1}-\sqrt{n})
=\sqrt{m}-1
$
so that
$\sum_{n=1}^{m-1} \dfrac{1}{\sqrt{n}}
\gt 2\sqrt{m}-2
$
so
$\sum_{n=1}^{m} \dfrac{1}{\sqrt{n}}
\gt 2\sqrt{m}-1+\frac1{\sqrt{m}}
\gt 2\sqrt{m}-1
$.
Therefore
$\sum_{n=1}^{m} \dfrac{1}{\sqrt{n}}$
is between
$2\sqrt{m}-1
$
and
$2\sqrt{m}+1
$.
A: Hint: You can show that $\sqrt{k+1}<\sqrt{k}+\frac{1}{\sqrt{k+1}}$.
