Prove that $\sum_{i=1}^n \frac{1}{\sqrt i} < 2 \sqrt n$ by induction - result doesn't match to wolfram alpha's result

Question

According to wolfram alpha, this inequality isn't true for $n > 2$. My result says otherwise. Assuming that the inequality is true for $n+1$,
$\sum_{i=1}^{n+1} \frac{1}{\sqrt i} < 2 \sqrt {n+1}$
by adding $\frac{1}{\sqrt {n+1}}$ to both sides of the original inequality, the following must be true $$ 2 \sqrt n + \frac{1}{\sqrt {n+1}} < 2 \sqrt{n+1}$$
the previous inequality is satisfied for $n\ge 0$. Have I done something wrong?

Answer 1

Note that $2\sqrt{n}+\frac{1}{\sqrt{n+1}}<2\sqrt{n+1}$ if and only if $2\sqrt{n(n+1)}<2(n+1)-1=2n+1$ if and only if $4n(n+1)<(2n+1)^2=4n^2+4n+1$ if and only if $0<1$. Therefore assuming $\sum_{i=1}^n\frac{1}{\sqrt{i}}<2\sqrt{n}$ gives $\sum_{i=1}^{n+1}\frac{1}{\sqrt{i}}<2\sqrt{n}+\frac{1}{\sqrt{n+1}}<2\sqrt{n+1}$ exactly as you suggested.

Also note that the function $f(x):=\frac{1}{\sqrt{x}}$ is strictly decreasing. Therefore $\sum_{i=1}^nf(i)=\sum_{i=0}^{n-1}((i+1)-i)f(i+1)<\int_0^nf(x)dx=2\sqrt{n}$ gives another proof of the inequality in question.

The wolfram output is indeed really weird.