I am trying to solve the following problem with mathematical induction but I can't quite seem to work it out.
Prove by mathematical induction that $$\sum_{r=1}^{n}{r^{-\frac12}}>n^{\frac 12}, \forall n\in \Bbb Z, n\ge 2$$
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2You should show what you have tried. Where specifically are you stuck? Have you done the base case at least? – Dave Jun 04 '18 at 18:11
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I am fine with the basis step, it is the inductive step that is not making sense; where I am required to show that the statement is true for the parameter value n = k + 1. – Steam Jun 04 '18 at 18:22
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For other ways than induction, the reference post is https://math.stackexchange.com/questions/2149448/other-idea-to-show-an-inequality-dfrac1-sqrt-1-dfrac1-sqrt-2-dfrac1/2149480#2149480 – zwim Jun 04 '18 at 18:52
3 Answers
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The important observation is that $$\sqrt{n+1}-\sqrt n=\frac1{\sqrt{n+1}+\sqrt n}\le \frac1{\sqrt {n+1}} $$
Hagen von Eitzen
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Multiply both sides by $\sqrt{n}$. Then you get $$ \sum_{r=1}^n \sqrt{\frac{n}{r}} \quad \mathrm{vs}\quad n $$ Now observe that in the left side each number is $\ge1$, and there are exactly $n$ terms. Since $n\geq2$ there is a number greater than 1 on the left side. So the sum is greater than $n$.
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@RhysHughes nothing, but you can always add a phrase "by the axiom of induction" to any part of the proof. This may not answer the question itself, but as you see, the proof is very easy, which is always a good thing to show. – Jun 04 '18 at 18:25
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Base Case: $r=2$
Show that $1^{-\frac12}+2^{-\frac12}>2^\frac12$, that is $1+\frac{1}{\sqrt{2}}>\sqrt{2}$
Then the inductive step: you should assume that for $n=k$ $$\sum_{r=1}^{k}{r^{-\frac 12}}>k^\frac12$$ and show that $k+1$ follows from that, i.e use it to infer: $$\sum_{r=1}^{k+1}{r^{-\frac 12}}>(k+1)^\frac12$$
As a hint: use that: $$\sum_{r=1}^{k+1}{r^{-\frac 12}}=\sum_{r=1}^{k}{r^{-\frac 12}}+(k+1)^{-\frac 12}$$
Rhys Hughes
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