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CHEGG Let k be any natural number. (a) Prove that for every positive integer n, we have σk(n) = σ−k(n)n k . Conclude that n is a perfect number exactly when σ−1(n) = 2. (b) Prove that for all positive integers n, we have σ1(n) ≤ n log(n + 1) + γn, where γ is Euler’s constant defined in class. (In this course, log x = loge x denotes the natural logarithm).

Respuesta :

Preciousorekha1

Answer:

Step-by-step explanation:

(a). from the question, we are asked to prove that for any positive n integer

σk(n) = σ−k(n)n∧k

taking K as a natural number

we have that

σ-k(n) = Σₐ₋ₙa∧k = Σₐ₋ₙ (n/a)∧-k

σ-k(n) = Σₐ₋ₙ (n/a)∧-k = Σₐ₋ₙ (a/n)∧k

  = Σₐ₋ₙ (a∧k/n∧k)

given that n∧k.σ-k(n) = Σₐ₋ₙ a∧k = σk(n)

prove : σk(n) = σ-k(n)*n∧k

(b). from the question, our positive integer is n

we have a convergent sequence  in an increasing order given as

(1 + 1/2 + 1/3 + --------+ 1/n - log (n+1)) ≥ 1

which is now thus

₹ = lim → ∞ (1 + 1/2 + 1/3 + --------+ 1/n - log (n+1))

∴ for all values of  n ≥1, 1 + 1/2 + 1/3 + --------+ 1/n - log (n+1) ≤ ₹

1 + 1/2 + 1/3 + --------+ 1/n ≤ ₹ + log (n+1) for all n≥ 1

recall that for natural numbers n in (a) i.e σ₁(n) : nσ₋₁(n)

σ₋₁(n) = Σₐ₋ₙ (1/a) ≤ Σn-i=1 (1) : ≤ ₹ + log(n + 1)

this gives σ₋₁(n) = nσ₋₁(n) ≤ n(log (n + 1) + ₹)

Prove : σ₋₁(n) = n(log (n + 1) + n₹ for value of n ≥ 1

cheers i hope this helps

akindelemf

Answer:

Step-by-step explanation:

Part b.  

Observe that the sequence ( 1+1/2+1/3+...+1/n - log(n+1) )n ≥1 is a monotonically increasing convergent sequence. Since a monotonically increasing sequence always converges to its supremum,

hence γ = [tex]\lim_{n \to \infty} (1+1/2+1/3+...+1/n - log(n+1))[/tex] = sup { 1+1/2+..+1/n - log(n+1) : n≥1 }

Therefore, for every n≥ 1, 1+1/2+1/3+..+1/n - log(n+1) ≤γ and this, 1+1/2+1/3+...+1/n ≤ γ + log(n+1) for all n≥ 1.

Now, from (a), δ₁(n) = nδ₋₁(n) for all natural numbers n.

Now, for all n ≥ 1,

σ-(n) = Σ/a «Σis y + log(n + 1) ζη + log(n+1) dn i=1

where the last inequality follows from the previous remarks.

Thus, (a), δ₁(n) = nδ₋₁(n) ≤ n(log(n+1) + γ ) = nlog (n+1) + nγ for all n ≥ 1.

This proves the inequality.

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