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I'm trying to solve the induction:

$ 1 + \frac{1}{\sqrt2} + ...+ \frac{1}{\sqrt{n}} > 2\sqrt{n+1} -2 $

My thought for solving this was:

$f(k) = 1 + \frac{1}{\sqrt2} + ...+ \frac{1}{\sqrt{k}}$

$f(k) + \frac{1}{\sqrt{k+1}} > 2\sqrt{k+2} -2$ - Need to proof

and I know that :

$f(k) + \frac{1}{\sqrt{k+1}} > 2\sqrt{k+1} -2 + \frac{1}{\sqrt{k+1}} $

So if I will try to show that :

$2\sqrt{k+1} -2 + \frac{1}{\sqrt{k+1}} > 2\sqrt{k+2} -2$

I manged to show that if I do square power to both sides of the equation (which keeps the inequality )the left side is bigger. I wanted to know if this way of thinking is good and valid, and if there are other ways to solve it, because I'm new to this concept.

Thanks

ə̷̶̸͇̘̜́̍͗̂̄︣͟
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Moran Tailu
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1 Answers1

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Basis for induction: $n=1$ gives LHS = $1$ and RHS = $0.828...$ so true

Get rid of the $2$ in your last inequality so LHS becomes $\frac{2k+3}{\sqrt{k+1}}$. This gives the inequality $$2k+3>2\sqrt{k^2+3k+2} \Leftarrow 4k^2 + 12k + 9 > 4k^2 +12k + 8$$ which is clearly true so the inequality $$1 + \frac{1}{\sqrt2} + ...+ \frac{1}{\sqrt{n}} > 2\sqrt{n+1} -2$$ is true for all $n \in \mathbb{N}$.

ə̷̶̸͇̘̜́̍͗̂̄︣͟
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