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Chapter 6: Problem 249

Prove by induction that \(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{n}} \geqslant \sqrt{n},\) for all \(n \geqslant 1\)

Short Answer

Expert verified
The inequality holds for all integers \(n \geq 1\) by induction.

Step by step solution

01

Base Case

First, let's verify the statement for the base case, when \(n = 1\).The left-hand side (LHS) of the inequality is:\[ \frac{1}{\sqrt{1}} = 1 \]The right-hand side (RHS) is:\[ \sqrt{1} = 1 \]Since \(1 \geq 1\), the base case holds true.
02

Inductive Hypothesis

Assume the inequality holds for some arbitrary positive integer \(k\). That is:\[ \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \cdots + \frac{1}{\sqrt{k}} \geq \sqrt{k} \]
03

Inductive Step

We need to prove that the inequality holds for \(k+1\).Consider:\[ \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \cdots + \frac{1}{\sqrt{k}} + \frac{1}{\sqrt{k+1}} \geq \sqrt{k+1} \]By the inductive hypothesis:\[ \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \cdots + \frac{1}{\sqrt{k}} \geq \sqrt{k} \]We aim to show:\[ \sqrt{k} + \frac{1}{\sqrt{k+1}} \geq \sqrt{k+1} \]
04

Prove Inductive Step

Squaring both sides of \( \sqrt{k} + \frac{1}{\sqrt{k+1}} \geq \sqrt{k+1} \) (to remove the square roots), we have:\[ (\sqrt{k} + \frac{1}{\sqrt{k+1}})^2 \geq k+1 \]Expanding the left side:\[ k + 2\frac{\sqrt{k}}{\sqrt{k+1}} + \frac{1}{k+1} \]The inequality becomes:\[ k + 2\frac{\sqrt{k}}{\sqrt{k+1}} + \frac{1}{k+1} \geq k + 1 \]Simplified, we get:\[ 2\frac{\sqrt{k}}{\sqrt{k+1}} + \frac{1}{k+1} \geq 1 \]Since \(2 \frac{\sqrt{k}}{\sqrt{k+1}} \geq 2 \frac{k}{k+1} \) and \(\frac{1}{k+1} \geq 0\), the inequality holds. Thus, the inductive step is proven.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Inequality Proof
An inequality proof requires showing that one expression is either less than or greater than another expression. To tackle such proofs, especially those appearing in mathematical induction, we often rely on algebraic manipulations and properties of numbers. The primary aim here is to demonstrate that the sum \[\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \cdots + \frac{1}{\sqrt{n}}\]agequates or outweighs \(\sqrt{n}\) for any integer \(n \geq 1\).
This involves checking both sides of inequality step-by-step to ensure their consistency across varying values of \(n\). Proofs of this nature help us understand behaviors of mathematical functions and sums as their variables grow large.

Remember that when you work with inequalities, each step must be justified either through logical reasoning or by algebraic proof that does not violate any prior assumptions. This step-by-step approach lays the groundwork for verifying inequalities in a rigorous manner.
Inductive Step
An inductive step forms the crux of mathematical induction. Once you confirm the base case, the next step involves assuming our statement is true for some arbitrary integer \(k\). The hallmark of induction is using this assumption to prove the proposition for \(k+1\).Here's how it works:
  • Start with the hypothesis: assume \[ \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \cdots + \frac{1}{\sqrt{k}} \geq \sqrt{k} \]
  • Expand this to show: \[ \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \cdots + \frac{1}{\sqrt{k+1}} \geq \sqrt{k+1} \]
  • Through mathematical manipulation and algebra, demonstrate the validity of this statement.

This logical process solidifies our belief in the pattern's correctness across different values through cumulative, verified steps. If the hypothesis works for \(k\) and you can show it works for \(k+1\), then by the principle of induction, it's true for all integer values \(n \geq 1\).
Base Case
In mathematical induction, the base case provides the initial confirmation of your proposition. It works by proving that the statement is true for the first logical value, usually \(n = 1\).For the given inequality:
  • Evaluate the left-hand side: \[ \frac{1}{\sqrt{1}} = 1 \]
  • Evaluate the right-hand side of the inequality: \[ \sqrt{1} = 1 \]
  • Since both sides equal and satisfy the inequality (\(1 \geq 1\)), the base case holds true.

This initial check is crucial as it anchors the inductive process, ensuring the statement holds from this foundational step and forwards through confirmed induction. Failure to establish the base case means the entire inductive proof could falter, being potentially incorrect for subsequent values.
Square Root Inequality
Square root inequalities involve expressions where the square root function plays a pivotal role, as observed in our inequality.To manage these:
  • Square both sides to simplify and clear the square roots, making expressions easier to handle.
  • In this example, we reshaped the inequality:\[\sqrt{k} + \frac{1}{\sqrt{k+1}} \geq \sqrt{k+1}\]
  • Expanding both sides, we reveal:\[(\sqrt{k} + \frac{1}{\sqrt{k+1}})^2 \geq k+1\]
  • Carefully expand and rearrange to keep the inequality intact, justifying each step algebraically.

By using methods such as squaring, we work past the square roots, ensuring the logic remains cinched tight across each manipulation. This foresight and care ensure that the conclusions we draw about the inequality stand firm under scrutiny.

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Most popular questions from this chapter

(a) Are 3,7,31,211 all prime? (b) Is \(2 \times 3 \times 5 \times 7 \times 11+1\) prime? (c) Is \(2 \times 3 \times 5 \times 7 \times 11 \times 13+1\) prime? We have already met two excellent historical examples of the dangers of plausible pattern-spotting in connection with Problem 118. There you proved that: "if \(2^{n}-1\) is prime, then \(n\) must be prime." You then showed that \(2^{2}-1,2^{3}-1,2^{5}-1,2^{7}-1\) are all prime, but that \(2^{11}-1=2047=23 \times 89\) is not. This underlines the need to avoid jumping to (possibly false) conclusions, and never to confuse a statement with its converse. In the same problem you showed that: "if \(a^{b}+1\) is to be prime and \(a \neq 1,\) then \(a\) must be even, and \(b\) must be a power of \(2 . "\) You then looked at the simplest family of such candidate primes namely the sequence of Fermat numbers \(f_{n}\) : $$ f_{0}=2^{1}+1=3, f_{1}=2^{2}+1=5, f_{2}=2^{4}+1=17, f_{3}=2^{8}+1=257, f_{4}=2^{16}+1 $$ It turned out that, although \(f_{0}, f_{1}, f_{2}, f_{3}, f_{4}\) are all prime, and although Fermat (1601-1665) claimed (in a letter to Marin Mersenne \((1588-1648))\) that all Fermat numbers \(f_{n}\) are prime, we have yet to discover a sixth Fermat prime! There are times when a mathematician may appear to guess a general result on the basis of what looks like very modest evidence (such as noticing that it appears to be true in a few small cases). Such "informed guesses" are almost always rooted in other experience, or in some unnoticed feature of the particular situation, or in some striking analogy: that is, an apparent pattern strikes a chord for reasons that go way beyond the mere numbers. However those with less experience need to realise that apparent patterns or trends are often no more than numerical accidents. Pell's equation (John Pell (1611-1685)) provides some dramatic examples. \- If we evaluate the expression " \(n^{2}+1\) " for \(n=1,2,3, \ldots,\) we may notice that the outputs \(2,5,10,17,26, \ldots\) never give a perfect square. And this is to be expected, since the next square after \(n^{2}\) is $$ (n+1)^{2}=n^{2}+2 n+1 $$ and this is always greater than \(n^{2}+1\). However, if we evaluate \(" 991 n^{2}+1 "\) for \(n=1,2,3, \ldots,\) we may observe that the outputs never seem to include a perfect square. But this time there is no obvious reason why this should be so - so we may anticipate that this is simply an accident of "small" numbers. And we should hesitate to change our view, even though this accident goes on happening for a very, very, very long time: the smallest value of \(n\) for which \(991 n^{2}+1\) gives rise to a perfect square is apparently

(Calkin-Wilf tree) The binary tree in the plane has a distinguished vertex as 'root', and is constructed inductively. The root is joined to two new vertices; and each new vertex is then joined to two further new vertices \(-\) with the construction process continuing for ever (Figure 11 ). Label the vertices of the binary tree with positive fractions as follows: \- the root is given the label \(\frac{1}{1}\) \- whenever we know the label \(\frac{i}{j}\) of a 'parent' vertex, we label its 'left descendant' as \(\frac{i}{i+j},\) and its 'right descendant' \(\frac{i+j}{j}\). (a) Prove that every positive rational \(\frac{r}{s}\) occurs once and only once as a label, and that it occurs in its lowest terms. (b) Prove that the labels are left-right symmetric in the sense that labels in corresponding left and right positions are reciprocals of each other. \(\triangle\)

(a) Assume the Least Element Principle. Suppose a subset \(T\) of the positive integers contains the integer "1", and that whenever \(k\) is in the set \(T\), then \(k+1\) is also in the set \(T\). Let \(S\) be the set of all positive integers which are not in the set \(T\). Conclude that \(S\) must be empty, and hence that \(T\) contains all positive integers. (b) Assume the Principle of Mathematical Induction. Let \(T\) be a non-empty set of positive integers, and suppose that \(T\) does not have a smallest element. Let \(S\) be the set of all positive integers which do not belong to the set \(T\). Prove that "1" must belong to \(S\), and that whenever the positive integer \(k\) belongs to \(S,\) then so does \(k+1 .\) Derive the contradiction that \(T\) must be empty, contrary to assumption. Conclude that \(T\) must in fact have a smallest element. \(\Delta\) To round off this final chapter you are invited to devise a rather different proof of the irrationality of \(\sqrt{2}\).

Let \(p\) be any prime number. Use induction to prove: \(" n^{p}-n\) is divisible by \(p\) for all \(n \geqslant 1\) ".

Prove by induction the statement: $$ " 1+3+5+\cdots+(2 n-1)=n^{2} \text { holds, for all } n \geqslant 1 " $$ The summation in Problem 231 was known to the ancient Greeks. The mystical Pythagorean tradition (which flourished in the centuries after Pythagoras) explored the character of integers through the 'spatial figures' which they formed. For example, if we arrange each successive integer as a new line of dots in the plane, then the sum " \(1+2+3+\cdots+n\) " can be seen to represent a triangular number. Similarly, if we arrange each odd number \(2 k-1\) in the sum " \(1+3+5+\cdots+(2 n-1)\) " as a " \(k\) -by- \(k\) reverse L-shape", or gnomon (a word which we still use to refer to the L-shaped piece that casts the shadow on a sundial), then the accumulated L-shapes build up an \(n\) by \(n\) square of dots \(-\) the "1" being the dot in the top left hand corner, the "3" being the reverse L-shape of 3 dots which make this initial "1" into a 2 by 2 square, the "5" being the reverse L-shape of 5 dots which makes this 2 by 2 square into a 3 by 3 square, and so on. Hence the sum " \(1+3+5+\cdots+(2 n-1)\) " can be seen to represent a square number. There is much to be said for such geometrical illustrations; but there is no escape from the fact that they hide behind an ellipsis (the three dots which we inserted in the sum between "5" and " \(2 n-1 "\), which were then summarised when arranging the reverse L-shapes by ending with the words "and so on"). Proof by mathematical induction, and its application in Problem 231, constitute a formal way of avoiding both the appeal to pictures, and the hidden ellipsis.
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