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

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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}

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Question ID: mtid-5-stid-8-sqid-2783-qpid-1340