Proving Induction: n∑(1/√n) > 2(√(n+1) -1)

[vela]

You know that
$$\sum_{k=1}^{n+1} \frac{1}{\sqrt k} > 2(\sqrt{n+1}-1) + \frac {1}{\sqrt{n+1}}.$$ Starting from there, you have to find a chain of true implications that eventually ends with
$$\sum_{k=1}^{n+1} \frac{1}{\sqrt k} > 2(\sqrt{n+2}-1).$$ A few posts back you said that you found
$$2(\sqrt{n+1}-1) + \frac {1}{\sqrt{n+1}} = \frac{2n+3-2\sqrt{n+1}}{\sqrt{n+1}}.$$ First, do a little algebra and show that this means
$$\sum_{k=1}^{n+1} \frac{1}{\sqrt k} > 2\left[\frac{n+\frac 32}{\sqrt{n+1}} - 1\right].$$ If you can now show that
$$2\left[\frac{n+\frac 32}{\sqrt{n+1}} - 1\right] > 2(\sqrt{n+2}-1),$$ you can use the transitive property of > to get the result you want. It's clear that this inequality holds if you can show that
$$\frac{n+\frac 32}{\sqrt{n+1}} > \sqrt{n+2}.$$ To prove this last inequality holds, you have to start with a statement that's undeniably true and go from there. What you can't do is start by assuming the inequality is true because that's what you're trying to prove.

What I suggest is that you start with
$$\sqrt{n+2} = \sqrt{n+2}\frac{\sqrt{n+1}}{\sqrt{n+1}}.$$ The goal is to use the rules of algebra to continue to manipulate the righthand side that so that's it's readily apparent that it's less than $\frac{n+\frac 32}{\sqrt{n+1}}$.

You've already done most of the work. It's just putting everything in the right order to make a valid logical argument.

 

[Ray Vickson]

So how do you show that the inequality is true?

The argument you gave above can be made to work: start with the fact that 8 < 9 and work backwards (going through your steps in the opposite order).

Alternatively, use the fact that $\sqrt{1+x} \leq 1 + x/2$ for $|x| < 1$, with strict inequality if $x \neq 0$. You can see this in two ways:
Method (1) look at the graph of $y = \sqrt{z}$; it lies below its tangent line through the point (1,1). The tangent line has slope 1/2, so we have $\sqrt{z} \leq 1 + (1/2)(z-1)$, with equality at $z = 1$ and a strict inequality otherwise.
Method (2): $1 +x \leq 1 + x + x^2/4 = (1 +\: x/2)^2$, so $\sqrt{1+x} \leq 1 + (x/2).$

Anyway, we have
\sqrt{n+2} = \sqrt{n+1}\sqrt{\left(1+\frac{1}{n+1} \right)}\\<br /> \leq \sqrt{n+1} \left( 1 + \frac{1}{2} \frac{1}{n+1} \right) = \sqrt{n+1}<br /> + \frac{1}{2\sqrt{n+1}}

 

[johann1301]

I have to get back to this problem at a later point, have some other things i have to go through before next lectures! But thank you all so much for your help!