Question: I'm trying to prove this by induction, but something doesn't add up. I see a solution given [here](https://www.algebra.com/algebra/homework/word/misc/Miscellaneous_Word_Problems.faq.question.29292.html), but it is actually proving that the expression is **greater** than $2\sqrt{n}$. I'd appreciate some insight.

Answer: 
Base step: 1<2.

Inductive step: $$\sum\_{j=1}^{n+1}\frac1{\sqrt{j}} < 2\sqrt{n}+\frac1{\sqrt{n+1}}$$
So if we prove
$$2\sqrt{n}+\frac1{\sqrt{n+1}}<2\sqrt{n+1}$$ 
we are done. Indeed, that holds true: just square the left hand side sides to get
$$4n+2\frac{\sqrt{n}}{\sqrt{n+1}}+\frac1{n+1}<4n+3<4n+4$$
which is the square of the right end side.

Errata: I forgot the double product in the square. The proof must be amended as follows:
$$2\sqrt{n}<2\sqrt{n+1}-\frac1{\sqrt{n+1}}$$ since by squaring it we get
$$4n<4n+4-4+\frac1{n+1}$$ which is trivially true.
Note that
$$ 2\sqrt{n+1}-2\sqrt n=2\cdot\frac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\sqrt{n+1}+\sqrt{n}}=2\cdot \frac{(n+1)-n}{\sqrt{n+1}+\sqrt{n}}<\frac 2{\sqrt n+\sqrt n}$$