/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 246 (a)(i) Explain why $$ \frac{... [FREE SOLUTION] | 91Ó°ÊÓ

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(a)(i) Explain why $$ \frac{1}{3^{2}}<\frac{1}{2^{2}} $$ So $$ \frac{1}{2^{2}}+\frac{1}{3^{2}}<\frac{2}{2^{2}}=\frac{1}{2} $$ (ii) Explain why \(\frac{1}{5^{2}}, \frac{1}{6^{2}}, \frac{1}{7^{2}}\) are all \(<\frac{1}{4^{2}},\) so $$ \frac{1}{4^{2}}+\frac{1}{5^{2}}+\frac{1}{6^{2}}+\frac{1}{7^{2}}<\frac{4}{4^{2}}=\frac{1}{4} $$ (b) Use part (a) to prove that $$ \frac{1}{1^{2}}+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\cdots+\frac{1}{n^{2}}<2, \text { for all } n \geqslant 1 $$ (c) Conclude that the endless sum $$ \frac{1}{1^{2}}+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\cdots+\frac{1}{n^{2}}+\cdots \text { (for ever) } $$ has a definite value, and that this value lies somewhere between \(\frac{17}{12}\) and 2 . $$ \Delta $$

Short Answer

Expert verified
The series sums to a definite value less than 2 and greater than \( \frac{17}{12} \).

Step by step solution

01

Compare Fractions

To show \( \frac{1}{3^2} < \frac{1}{2^2} \), calculate each square. Here, \( 3^2 = 9 \) and \( 2^2 = 4 \). So, \( \frac{1}{9} < \frac{1}{4} \) since a larger denominator means a smaller fraction.
02

Add and Compare

Adding the fractions \( \frac{1}{2^2} \) and \( \frac{1}{3^2} \), which are \( \frac{1}{4} + \frac{1}{9} = \frac{9 + 4}{36} = \frac{13}{36} \). Clearly, \( \frac{13}{36} < \frac{1}{2} \), because \( \frac{1}{2} = \frac{18}{36} \).
03

Further Fraction Comparison

For \( \frac{1}{5^2}, \frac{1}{6^2}, \frac{1}{7^2} \) compared to \( \frac{1}{4^2} \), calculate the squares: \( 25, 36, 49 \, \) and \( 16 \). So, \( \frac{1}{25}, \frac{1}{36}, \frac{1}{49} < \frac{1}{16} \).
04

Sum and Compare

Summing \( \frac{1}{4^2} + \frac{1}{5^2} + \frac{1}{6^2} + \frac{1}{7^2} = \frac{1}{16} + \frac{1}{25} + \frac{1}{36} + \frac{1}{49} \). Add these and notice they total less than \( \frac{4}{16} = \frac{1}{4} \).
05

Prove the Series is Bounded (Part b)

Using sums: \( \frac{1}{1^2} = 1 \), and parts (a)(i) and (a)(ii), for any \( n > 3 \), similar smaller fractions total less than corresponding \( 1/k \), implying sum less than 2.
06

Conclude Definiteness and Bounds (Part c)

Given each finite partial sum is bounded and smaller additions correspondingly decrease to a limit, the series is bounded by 2. Therefore, the infinite sum has a definite value. Since smaller portions accumulate, the value lies between \( \frac{17}{12} \) and 2.

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Key Concepts

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

Convergence
Convergence in terms of infinite series refers to the behavior where the sum of an infinite sequence of numbers approaches a specific value as more and more terms are added. It's like walking closer to a fixed point. In mathematical terms, for a series to converge, the partial sums of the series must get closer and closer to a definite value. If it does, the series is called convergent.

In our exercise, the series involves terms like \( \frac{1}{n^2} \). As \( n \) becomes very large, each term \( \frac{1}{n^2} \) becomes very small, essentially shrinking towards zero. This shrinking effect is crucial, as it suggests that adding such small values continually results in a sum that settles near a certain number. This is the essence of convergence in our context. For the given series, this specific target value or sum lies between certain bounds, such as \( \frac{17}{12} \) and 2, indicating its convergent nature.
Comparison of Fractions
Comparison of fractions is foundational in understanding inequalities within infinite series. When comparing fractions like \( \frac{1}{3^2} \) and \( \frac{1}{2^2} \), think of the denominators. A larger denominator in a fraction results in a smaller value. Thus, \( \frac{1}{9} < \frac{1}{4} \) because 9 is larger than 4.

These comparisons are useful when establishing bounds for sums in the bigger picture of convergence. By comparing fractions, we essentially organize and estimate the behavior of terms in a series, which helps in demonstrating convergence or bounding the sum. For instance, comparing multiple terms to a common denominator allows us to sum and evaluate parts of the series effectively, reinforcing our understanding of where the series converges.
Bounding Series
Bounding a series involves defining numerical limits within which the series' sums fall. Establishing bounds ensures the series doesn't grow without limit, showcasing its convergent nature. Essentially, bounding is about framing the sum so it doesn't surpass certain values. In our example, specific parts of the series like \( \frac{1}{4^2} + \frac{1}{5^2} + ... \), contribute less than a particular fraction, ensuring the overall sum remains stable.

Understanding these bounds helps demonstrate the convergence process, as it provides a way to predict the limit towards which the series sums are heading. These estimated bounds offer the theoretical backing needed to substantiate claims about the series' convergence properties and its ultimate sum.
Partial Sums
Partial sums are the summation of a select number of initial terms in a series. Think of them as snapshots of the series as it builds towards an infinite sum. These are calculated by adding the terms one by one, starting from the first and moving forward.

In our series, the partial sum calculation allows one to see the gradual growth of the sum. For instance, summing only the first few terms, like \( \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} \), gives a glimpse into how the series begins to shape up and converge.

The concept of partial sums is crucial in proving convergence because it shows how, even as more terms are added, the sum approaches a specific value — providing evidence that the series stabilizes within a predictable range. This methodical approach allows mathematicians to reach conclusions about the series' overall behavior and final sum.

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Most popular questions from this chapter

Prove the statement: \(" 5^{2 n+2}-24 n-25\) is divisible by \(576,\) for all \(n \geqslant 1 "\) When trying to construct proofs in private, one is free to write anything that helps as 'rough work'. But the intended thrust of Problem 229 is two-fold: \- to introduce the habit of distinguishing clearly between (i) the statement \(\mathbf{P}(n)\) for a particular \(n,\) and (ii) the statement to be proved \(-\) namely \(" \mathbf{P}(n)\) is true, for all \(n \geqslant 1 " ;\) and to draw attention to the "induction step" (i.e. the third bullet point above), where (i) we assume that some unspecified \(\mathbf{P}(k)\) is known to be true, and (ii) seek to prove that the next statement \(\mathbf{P}(k+1)\) must then be true. The central lesson in completing the "induction step" is to recognize that: to prove that \(\mathbf{P}(k+1)\) is true, one has to start by looking at what \(\mathbf{P}(k+1)\) says.

Let \(f_{1}=2, f_{k+1}=f_{k}\left(f_{k}+1\right)\). Prove by induction that \(f_{k}\) has at least \(k\) distinct prime factors.

(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

(a) Let $$ s_{n}=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\cdots \pm \frac{1}{n} $$ (where the final operation is " \(+"\) if \(n\) is odd and " \(-"\) if \(n\) is even). (i) Prove that $$ s_{2 n-2}

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 \(n \geqslant 1\)

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