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For any natural number n, prove the validity of given series by mathematical induction:

$2(\sqrt{n+1}-1)<1+(\dfrac{1}{\sqrt{2}})+...+(\dfrac{1}{\sqrt{n}})<2\sqrt{n}$

Let $P(n)$ be the proposition that for the first $n$ positive integers

$2(\sqrt{n+1}-1)<1+(\dfrac{1}{\sqrt{2}})+...+(\dfrac{1}{\sqrt{n}})<2\sqrt{n}$

BASIS STEP: $P(1)$ is true, because

$2(\sqrt{1+1}-1)<1<2\sqrt{1}$

$2(\sqrt{2}-1)<1<2$

INDUCTIVE STEP: For the inductive hypothesis we assume that $P(k)$ holds for an arbitrary positive integer $k.$ That is, we assume that

$2(\sqrt{k+1}-1)<1+(\dfrac{1}{\sqrt{2}})+...+(\dfrac{1}{\sqrt{k}})<2\sqrt{k}$

Under this assumption, it must be shown that $P(k+1)$ is true, namely, that

$2(\sqrt{(k+1)+1}-1)<1+(\dfrac{1}{\sqrt{2}})+...+(\dfrac{1}{\sqrt{k}})$

$+(\dfrac{1}{\sqrt{k+1}})<2\sqrt{k+1}$

We have that

$2(\sqrt{k+1}-1)+(\dfrac{1}{\sqrt{k+1}})<1+(\dfrac{1}{\sqrt{2}})+...+(\dfrac{1}{\sqrt{k}})+(\dfrac{1}{\sqrt{k+1}})$

$<2\sqrt{k}+(\dfrac{1}{\sqrt{k+1}})$

Show that

$2(\sqrt{k+2}-1)<2(\sqrt{k+1}-1)+(\dfrac{1}{\sqrt{k+1}})$

$2(\sqrt{k+2}-\sqrt{k+1})<\dfrac{1}{\sqrt{k+1}}$

$\dfrac{2(k+2-k-1)}{\sqrt{k+1}+\sqrt{k+2}}<\dfrac{1}{\sqrt{k+1}}$

$\dfrac{2}{\sqrt{k+1}+\sqrt{k+2}}<\dfrac{1}{\sqrt{k+1}}$

$\dfrac{2}{\sqrt{k+1}+\sqrt{k+2}}<\dfrac{2}{\sqrt{k+1}+\sqrt{k+1}}=\dfrac{1}{\sqrt{k+1}},$

True for $k\geq1.$

Show that

$2\sqrt{k}+(\dfrac{1}{\sqrt{k+1}})<2\sqrt{k+1}$

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

$\dfrac{2}{\sqrt{k}+\sqrt{k+1}}>\dfrac{1}{\sqrt{k+1}}$

$\dfrac{2}{\sqrt{k}+\sqrt{k+1}}>\dfrac{2}{\sqrt{k+1}+\sqrt{k+1}}=\dfrac{1}{\sqrt{k+1}}$

True for $k\geq1.$

We show that $P(k+1)$ is true under the assumption that $P(k)$ is true. This completes the inductive step.

We have completed the basis step and the inductive step, so by mathematical induction we know that $P(n)$ is true for all positive integers $n.$

That is, we have proven that for any natural number $n,$

$2(\sqrt{n+1}-1)<1+(\dfrac{1}{\sqrt{2}})+...+(\dfrac{1}{\sqrt{n}})<2\sqrt{n}$