Almost Harmonic

Calculus Level 3

The harmonic series is the sum of the reciprocals of all natural numbers; mathematically, ∑ i = 1 ∞ 1 i . \displaystyle\sum_{i=1}^\infty\dfrac{1}{i}. This series has an infinite sum, as first proved by Nicole Oresme in the 14th century.

Find ⌊ 1000 N ⌋ , \lfloor1000N\rfloor, where N N is the largest possible non-negative K K for which the following sum does not \textbf{not} diverge to − ∞ . -\infty. ∑ i = 1 ∞ 1 i − K \sum_{i=1}^\infty\dfrac{1}{i}-K


The answer is 0.

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Trevor B. by Trevor B. , 22, USA

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1 solution

Trevor B.
Apr 6, 2014

Since the harmonic series has an infinite sum, this means that it is greater than any rational number. We will prove that no positive K K exists.

Let K K be an arbitrary number in the range ( 0 , 1 ) (0,1) such that the condition holds. There are a finite number of terms in the sequence greater than K K and an infinite number of terms less than K . K. Since a finite number of finite terms added together is finite, the sum of the terms greater than K K is finite. However, since the sum of the whole thing is infinite, the sum of the smaller than K K must be infinite. Subtracting K K from those terms will cause a repeated adding of negative terms forever. No matter how small K K is, as the terms of the harmonic sequence get closer and closer to 0 , 0, the terms minus K K will go to − K . -K. This will occur an infinite number of times. Since K K is positive, ∞ × − K = − ∞ , \infty\times-K=-\infty, which contradicts the original condition. Thus, there is no positive K K that satisfies the condition.

It is trivial to see that K = 0 K=0 satisfies the conditions, making the sequence diverge to + ∞ , +\infty, so N = 0 N=0 and ⌊ 1000 N ⌋ = 0 \lfloor1000N\rfloor=\boxed{0}

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Alternate solution

WARNING: This section contains calculus. If you solved this problem using the above solution, good job. If you do not know calculus, then don't read this.

If there exists a value of K K in the problem that makes the summation equal a real number, then there exists values of K K that will make the sum approach every real number in the range ( − ∞ , ∞ ) (-\infty,\infty) (you can prove this using the rules of continuity). Let's try to home in on one such value: 0. 0. The value of K K that will produce 0 0 is the average of all of the terms in the harmonic series.

One of the proofs that the harmonic series has an infinite sum uses this fact. lim ⁡ n → ∞ ∫ 1 n 1 x d x = lim ⁡ n → ∞ ln ⁡ x ∣ 1 n = lim ⁡ n → ∞ ln ⁡ n − ln ⁡ 1 = ∞ \lim_{n\rightarrow\infty}\int_1^n\dfrac{1}{x}\text{ }dx=\lim_{n\rightarrow\infty}\left.\ln x\right|^n_1=\lim_{n\rightarrow\infty}\ln n-\ln1=\infty We can try to find the average of all of the terms in the harmonic series by applying the average value formula on that integral. Let's get it to something easy to work with first. lim ⁡ n → ∞ 1 n − 1 ∫ 1 n 1 x d x = lim ⁡ n → ∞ 1 n ∫ 1 n 1 x d x \lim_{n\rightarrow\infty}\dfrac{1}{n-1}\int_1^n\dfrac{1}{x}\text{ }dx=\lim_{n\rightarrow\infty}\dfrac{1}{n}\int_1^n\dfrac{1}{x}\text{ }dx We can change the denominator before the integral because it will diverge to ∞ \infty no matter what the constant being added to it is.

Now let's solve the limit.

lim ⁡ n → ∞ 1 n ∫ 1 n 1 x d x = lim ⁡ n → ∞ ∫ 1 n 1 n x d x = lim ⁡ n → ∞ ln ⁡ x n ∣ 1 n = lim ⁡ n → ∞ ln ⁡ n − ln ⁡ 1 n = lim ⁡ n → ∞ ln ⁡ n n = lim ⁡ n → ∞ ln ⁡ n n \begin{array}{c}\lim_{n\rightarrow\infty}\dfrac{1}{n}\int_1^n\dfrac{1}{x}\text{ }dx&=\lim_{n\rightarrow\infty}\int_1^n\dfrac{1}{nx}\text{ }dx\\ &=\lim_{n\rightarrow\infty}\left.\dfrac{\ln x}{n}\right|^n_1\\ &=\lim_{n\rightarrow\infty}\dfrac{\ln n-\ln 1}{n}\\ &=\lim_{n\rightarrow\infty}\dfrac{\ln n}{n}=\lim_{n\rightarrow\infty}\ln\sqrt[n]{n} \end{array}

Since lim ⁡ n → ∞ n n = 1 , \lim_{n\rightarrow\infty}\sqrt[n]{n}=1, lim ⁡ n → ∞ ln ⁡ n n = 0. \lim_{n\rightarrow\infty}\ln\sqrt[n]{n}=0.

So it appears that the average value of 1 x \frac{1}{x} from 1 1 to n n as n n grows larger will approach 0. 0. This makes our K K approach 0. 0. You can prove this for all other real numbers we try to converge the sequence in question to, so every value of K K we try will equal 0. 0. Therefore, N = 0 , N=0, and ⌊ 1000 N ⌋ = 0 . \lfloor1000N\rfloor=\boxed{0}.

Trevor B. - 7 years, 2 months ago

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To the person who changed this from Algebra to Calculus: I think it's more Algebra because you have to take note of the fact that the sum of the terms of the harmonic sequence less than K K is infinite. This was a solution involving calculus, but there are multiple solutions to every problem.

Trevor B. - 7 years, 2 months ago

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