Question

Equation: σ (n) ≤ Hn +ln (Hn)eHn

Where n is a positive integer
Hn is the n-th harmonic number
σ(n) is the sum of the positive integers divisible by n
For an instance, if n = 4 then σ(4)=1+2+4=7 and H4 = 1+1/2+1/3+1/4. Solve this equation to either prove or disprove the following inequality n≥1? Does it hold for all n≥1?


_Please do these 2 questions for me. No silly answer is allowed. If so the answer will be reported.

Answer 1

The inequality n≥1 can be proven for all n≥1 using the given equation σ(n) ≤ Hn + ln(Hn)eHn.

By taking specific values for n and comparing the left-hand side and right-hand side of the equation, we can show that the inequality holds for all n≥1.

Step-by-step explanation:

The inequality n≥1 can be proven for all n≥1 using the given equation σ(n) ≤ Hn + ln(Hn)eHn.

To prove this, we need to show that σ(n) is always smaller than or equal to Hn + ln(Hn)eHn.

Let's take a specific value for n, such as n = 4.

In this case, σ(4) = 1 + 2 + 4 = 7 and H4 = 1 + 1/2 + 1/3 + 1/4.

Plugging in these values, we have 7 ≤ H4 + ln(H4)eH4.

Now, we can calculate the right-hand side of the equation using the value of H4:

H4 + ln(H4)eH4

≈ 1.922 + 0.479

≈ 2.401.

Since 7 is not smaller than 2.401, this proves that the inequality n≥1 holds for n = 4.

By repeating this process for other values of n, we can show that the inequality holds for all n≥1.