Show that Σ1/c= ∞ where c ranges over all composite numbers.(b)ShowΣ(from i=2 to i=∞)1/logi=∞(c)showΣ1/ilogi=∞

That is show that


a) Σ1/c= ∞


that is the sum of 1/c equals infintie (∞) where c ranges over all composite numbers.





b) Σ(from i=2 to i=∞) 1/log i = ∞


that is the sum of 1/log i equals infinty where i ranges from 2 to inifity ∞





c) show that Σ (from i=2 to i=∞) 1/ i log i = ∞


that is the sum of 1/ i log i ( 1 divided by i times log i) equals infinity ∞ where i ranges from 2 to infinity ∞


(hint is that one can divide this into blocks)








d) show that Σ (from i=2 to i=∞) squared(1/ i log i) = finite number





that is sum of the square of 1/ i log i is finite .





that is (1/i log i) (1/ i log i) is a finite number.








e ) lastly show that for every € %26gt; 0 there is an integer n so that π(n) / n %26lt; € .


( you can use the facts about φ(n) )





note: π(n) is the number of positive divisors of n i think


φ(n) is is the euler phi function and is defined as the number of integers n for 1%26lt;n%26lt;asuch that gcd (a,n) =

Show that Σ1/c= ∞ where c ranges over all composite numbers.(b)ShowΣ(from i=2 to i=∞)1/logi=∞(c)showΣ1/ilogi=∞
a: Since all multiples of 2 except for 2 itself are composite, we have [c is composite]∑1/c ≥ [n=2, ∞]∑1/(2n) = 1/2 * [n=2, ∞]∑1/n = 1/2*∞ = ∞





b: Note that for all n, ln n %26lt; n, so 1/ln n %26gt; 1/n. Since [n=2, ∞]∑1/n diverges, so does [n=2, ∞]∑1/ln n by comparison





c: By the integral test, we have that [n=2, ∞]∑1/(n ln n) converges if and only if [2, ∞]∫1/(x ln x) dx converges. But [2, ∞]∫1/(x ln x) dx = ln ln x |[2, ∞] = ln ln ∞ - ln ln 2 = ∞, so 1/(n ln n) does not converge





d: By comparison. Note that for every n≥3, 1/(n ln n)² %26lt; 1/n², so if [n=2, ∞]∑1/n² converges, then so does [n=2, ∞]∑1/(n ln n)². By the integral test, [n=2, ∞]∑1/n² converges iff [2,∞]∫1/x² dx converges. Since [2, ∞]∫1/x² dx = -1/x |[2, ∞] = -1/∞ + 1/2 = 1/2, the integral converges, thus so do both sums.





e: I don't think your problem statement is right -- the notation π(n) refers to the prime counting function, which is the number of prime numbers less than or equal to a particular number.





In any case, the proof relies on noting that every prime number less than n is also coprime to it, so for every n, π(n) ≤ φ(n) (note that this does not have to be weakened to π(n) ≤ φ(n) + 1, since while n itself may be a prime number which is less than or equal to n and yet not coprime to n, there is always one number coprime to n which is not n itself, namely 1, so that takes its place). Therefore, π(n)/n ≤ φ(n)/n, and it suffices to show that for every ε%26gt;0 ∃n s.t. φ(n)/n %26lt; ε.





To show this, we note that





[n=1, ∞]∑1/n = [p prime]∏([k=0, ∞]∑1/p^k)





This follows from the fact that if you take the Cauchy product of all the infinite series on the right, you will get exactly one term of the form 1/(p₁^k₁*p₂^k₂*p₃^k₃... p_n^k_n) for every finite tuple of prime numbers p₁, p₂.... p_n and nonnegative integers k₁, k₂... k_n. In other words, you will have exactly one term for each possible prime factorization, and since the fundamental theorem of arithmetic establishes that there is a one-to-one correspondence between positive integers and prime factorizations, there is exactly one term of the form 1/n for each positive integer n, so the sum is [n=1, ∞]∑1/n.





Now, simplifying the infinite product:





[p prime]∏([k=0, ∞]∑1/p^k) = [p prime]∏1/(1 - 1/p)





Which follows from the well-known formula for the sum of a geometric series. So combining these two equalities, we have:





[p prime]∏1/(1 - 1/p) = [n=1, ∞]∑1/n





Since [n=1, ∞]∑1/n diverges, it follows that [p prime]∏1/(1 - 1/p) diverges to infinity -- that is, that for every natural number M, there exists N such that the product of 1/(1- 1/p) over all primes less than N is greater than M. Now, let ε%26gt;0 be arbitrary, then choose M such that 1/M %26lt; ε and choose N large enough so that [p prime, p%26lt;N]∏1/(1-1/p) %26gt; M, then taking reciprocals, [p prime, p%26lt;N]∏(1-1/p) %26lt; ε. So if we can find n such that φ(n)/n = [p prime, p%26lt;N]∏(1-1/p), then we are done. But that is easy, for just let n = [p prime, p%26lt;N]∏p, then since all the prime factors of n are distinct, we have:





φ(n) = [p | n]∏(p - 1) = [p prime, p%26lt;N]∏(p-1)





And thus:





φ(n)/n = ([p prime, p%26lt;N]∏(p-1))/([p prime, p%26lt;N]∏p)


= [p prime, p%26lt;N]∏((p-1)/p) = [p prime, p%26lt;N]∏(1 - 1/p) %26lt; ε





And so π(n)/n %26lt; φ(n)/n %26lt; ε, and we are done.