Prove by induction that
$$1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+…+\frac{1}{\sqrt{n}}<\sqrt{n}, (n\ge2)$$
Induction basis is correct: $$\frac{1}{\sqrt{2}}<\sqrt{2}$$
So we need to prove that
$$1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+…+\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}<\sqrt{n+1}$$
One of my attempts was to replace $1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+…+\frac{1}{\sqrt{n}}$ by a number that we know from the induction hypothesis to be greater than it, which is $\sqrt{n}$, and if we show that the inequality is still true then we prove the initial statement.
$$\sqrt{n}+\frac{1}{\sqrt{n+1}}\overset{?}{<}\sqrt{n+1}$$
And, unfortunately, the inequality doesn't hold. I checked it in Wolfram Alpha and the difference between one side of the inequality and the other is really really small. In my opinion, that is why this inequality is so tough to prove (for me, obviously).
Thanks for your help.
Best Answer
The inequality is not true: $$ 1+\frac1{\sqrt2}\simeq1.707>1.4142\simeq\sqrt2. $$ The reverse inequality can be proved by induction, as you are trying: the inductive step would be $$ \sqrt n+\frac1{\sqrt{n+1}}>\sqrt{n+1}. $$ This can be seen as follows: from $n+1>n$, we get $$ n(n+1)>n^2. $$ Taking square root and adding $1$, $$ \sqrt{n(n+1)}+1>n+1. $$ Now divide by $\sqrt{n+1}$: $$ \sqrt{n}+\frac1{\sqrt{n+1}}>\sqrt{n+1}. $$