Problem Sheet 1. 1) Use Theorem 1.1 to prove that. 1 p loglogx 1

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1 Problem Sheet ) Use Theorem. to prove that p loglog for all real 3. This is a version of Theorem. with the integer N replaced by the real. Hint Given 3 let N = [], the largest integer. Then, importantly, for a sum over any terms a n, n a n = n N a n. ) Show that for every integer k there eist a consecutive sequence of k composite integers i.e. non-primes). 3) In the proof of Theorem. it is shown that for integer N. Prove that for all real. logn +) log n N n n logn +, n log+ 4) Let f be a function integrable on [,N]. Prove that for integers N >, i) if f is increasing f t)dt+f ) ii) if f is decreasing n N f t)dt+f N) n N f n) f n) f t)dt+f N), f t)dt+f ).

2 Hint Apply 6). These two parts can be summed up by saying that if f is monotonic then minf ),f N)) f n) for all integers N >. n N f t)dt maf ),f N)), 5) Prove that for integers N N logn N + log educe that for such N, n N ) N N N! < N logn N +)logn N + log. ) ) N N. e Hint It might help to note that n N logn = n N logn. 6) i) Prove that for θ > 0, θ N θ n N for all integers N >. n N θ θ θ θ θ, ii) You cannot put θ = into part i because of the θ in the denominator but what is the limit N θ θ lim? θ θ 7) Show that ) p log, 6) for all real 3. Thus deduce that lim ) = 0. p 3

3 Hint Replace by an integer, look at the inverse of the product, and use ideas and results from the proof of Theorem.. Aside Later in the course we will show, for Level 4 students, that with an appropriate constant c we have ) c p log as. Notation Here f ) g) as means lim f )/g) =. 8) i) Use 6) to show that logy < y for all real y >. ii) By an appropriate choice of y in part i show that for all n we have for >. log < n /n iii) educe that for all ε > 0, we have log ε ε. This means that the logarithm of grows slower than any power of. Notation Here f ) ε g) means that there eists a constant C = Cε), depending on ε, such that f ) < Cε)g). 9) An important task throughout this course is the estimating of integrals. The first is logy) l, with integer l. An application of 6) gives log) l logy) l log) l. 7) i) Which of these bounds do you think represents more truly the rate of growth of the integral as? Hint Integrate by parts. ii) Apply a tweak and split the integral at : logy) = l logy) + l 4 logy) l.

4 Apply the upper bound in 6) to both parts to show logy) l l log) l. 8) Hint You may need to make use of Question 8 in the integral over [, ]. iii) Show, using integration by parts a number of times and then Part ii, that for m we have m ) logy) = l +j)! l l )! log l+j +O l,m log l+m+ j=0 Notation O l,m /log l+m+ ) represents a function of,f ) say, about which all we know is that f ) l,m /log l+m+, and for this notation see Question 8! 0) Another integral we will meet in the course is the tail end i.e. from to ) logy) l, 9) y with integer l. i) Use log < l /l), which follows from Question 8, to show that logy) l y l)l /. o you think this upper bound truly represents the size of the integral? ii) We cannot directly apply 6) to 9) because the interval of integration is infinite. Instead tweak as in Question 9 but this time split the integral at : logy) l logy) l l logy) = +. y y y Now remove the log factors from both integrals, leaving an integration over a power of y in both cases. Thus show that logy) l log) l y l. 5

5 ) Let π) = be the number of primes less than equal to. The infinitude of primes is equivalent to lim π) = Prove that π) clog for some constant c > 0. Justify each step in the following argument. With > a constant to be chosen π) log < log log e log. log < p log p n log loglog log n )) log 6