Induction: Bounds on sum of inverses of first $n$ square roots I am trying to show by induction that $2(\sqrt{n}-1)<1+\dfrac{1}{\sqrt{2}}+\cdots+\dfrac{1}{\sqrt{n}}<2\sqrt{n}$.
For the upper bound, I have that if $1+\dfrac{1}{\sqrt{2}}+\cdots+\dfrac{1}{\sqrt{n}}<2\sqrt{n}$, then $1+\dfrac{1}{\sqrt{2}}+\cdots+\dfrac{1}{\sqrt{n}}+\dfrac{1}{\sqrt{n+1}}<2\sqrt{n}+\dfrac{1}{\sqrt{n+1}}=$
$2\sqrt{n}+\dfrac{2}{2\sqrt{n+1}}<2\sqrt{n}+\dfrac{2}{\sqrt{n}+\sqrt{n+1}}=2\sqrt{n}+2(\sqrt{n+1}-\sqrt{n})=2\sqrt{n+1}$.
However, despite hours of trying, I cannot figure out how to show that $2(\sqrt{n+1}-1)<1+\dfrac{1}{\sqrt{2}}+\cdots+\dfrac{1}{\sqrt{n}}+\dfrac{1}{\sqrt{n+1}}$.

Try to use the induction method for $2(\sqrt{n}-1+\frac{1}{\sqrt{n}}) < \frac{1}{\sqrt{1}} +  \frac{1}{\sqrt{2}} + \cdots + \frac{1}{\sqrt{n}}$
Check it for base cases like $n=1,2,3,4,5,6,7,8,9,10$ and it will fail for all $n=1,2,3,4,5,6,7$ and will hold for $n=8,9,10$.
Induction step : assume that $2(\sqrt{n}-1+\frac{1}{\sqrt{n}}) < \frac{1}{\sqrt{1}} +  \frac{1}{\sqrt{2}} + \cdots + \frac{1}{\sqrt{n}}$ is true for the number $n$.
Proving step : we need to prove that $2(\sqrt{n+1}-1+\frac{1}{\sqrt{n+1}}) < \frac{1}{\sqrt{1}} +  \frac{1}{\sqrt{2}} + \cdots + \frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}$
From the induction step we know that $2(\sqrt{n}-1+\frac{1}{\sqrt{n}}) < \frac{1}{\sqrt{1}} +  \frac{1}{\sqrt{2}} + \cdots + \frac{1}{\sqrt{n}}$ thus replacing $ \frac{1}{\sqrt{1}} +  \frac{1}{\sqrt{2}} + \cdots + \frac{1}{\sqrt{n}}$ with $2(\sqrt{n}-1+\frac{1}{\sqrt{n}})$ will strengthen the inequality.
We arrive at $2(\sqrt{n+1}-1+\frac{1}{\sqrt{n+1}}) < 2(\sqrt{n}-1+\frac{1}{\sqrt{n}})+\frac{1}{\sqrt{n+1}}$, (i will leave this for you to prove).
Now since its obvious that $2(\sqrt{n}-1+\frac{1}{\sqrt{n}}) > 2(\sqrt{n}-1)$ then the inequality $2(\sqrt{n}-1) < \frac{1}{\sqrt{1}} +  \frac{1}{\sqrt{2}} + \cdots + \frac{1}{\sqrt{n}}$ will be a corollary to this induction proof.
you just have to check basic cases for $2(\sqrt{n}-1) < \frac{1}{\sqrt{1}} +  \frac{1}{\sqrt{2}} + \cdots + \frac{1}{\sqrt{n}}$ which are $n=1,2,3,4,5,6,7$ and thus completing the proof.
