Mistakes on Brilliant

Calculus Level 4

Let M M be the set of positive integers ending at 2018 2018 , and M C M^{\small C} be the set of positive integers not ending at 2018 2018 . Then,

True or False?

  • a) ∑ m ∈ M 1 m \displaystyle \sum_{m \in M} \frac{1}{m} converges.

  • b) ∑ m ∈ M C 1 m \displaystyle \sum_{m \in M^{\small C}} \frac{1}{m} converges.

Bonus.- Generalize for any positive integer

a)True, b) True a)False b)False a)False b)True a)True, b) False

Guillermo Templado by Guillermo Templado , 46, Spain

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1 solution

a) The partial sums of ∑ m ∈ M 1 m = ∑ n = 0 ∞ 1 10000 ⋅ n + 2018 \displaystyle \sum_{m \in M} \frac{1}{m} = \sum_{n = 0}^{\infty} \frac{1}{10000 \cdot n + 2018} are greater than partial sums of ∑ m ∈ Z + 1 20000 ⋅ m = 1 20000 ∑ m ∈ Z + 1 m → ∞ , as m → ∞ \displaystyle \sum_{m \in \mathbb{Z}^{+}} \frac{1}{20000 \cdot m} = \frac{1}{20000} \sum_{m \in \mathbb{Z}^{+}} \frac{1}{ m} \to \infty, \space \text{ as } m \to \infty because harmonic series diverges, so the first sum diverges.

b) Let P \mathfrak{P} be the set of prime numbers, then ∑ m ∈ M C 1 m ≥ ∑ p ∈ P 1 p \displaystyle \sum_{m \in M^{\small C}} \frac{1}{m} \ge \sum_{p \in \mathfrak{P}} \frac{1}{p} and due to this last series diverges, the first one too.

Note.- If someone wishes a proof for sum of inverses of prime numbers, he can ask me, and I'll do it.

Bonus.- The same behaviour happens with other any positive integer.

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