9
$\begingroup$

Consider the sum $$S = \frac{1}{18} + \frac{1}{19} + \cdots + \frac{1}{47}$$. A brute-force calculation (okay, I just used Wolfram Alpha) shows that $$ S = \frac{442017301628992345493}{442720643463713815200} < 1$$

My question is whether there is some way to recognize that this sum is less than 1 without brute-force methods. For example, is there some way to pair or group the terms together so that one can say something like "These terms sum to less than 1/6, these terms sum to less than 1/6, etc., so the whole thing sums to less than 1"?

(Note: if this problem looks familiar, it may be because it is a companion to Is there an easy way to see that ${1\over5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{11} + \frac{1}{12} > 1$?. However, in that problem, the goal was to show that a particular sum of unit fractions was greater than 1, and the answers there used the fact that $f(x) = \frac{1}{x}$ is convex to argue that certain combinations are greater than specific values. It's unclear to me whether those strategies can be adapted to show that a combination is less than a target value.)

Edited to add: I guess I should clarify that anything requiring the calculation of a logarithm or using an approximate value of $e$ would fall, in my opinion, under the heading of "brute force". I'm looking for things that can be done without the use of a calculator.

$\endgroup$
2
  • $\begingroup$ The difference from 1 is less than 1/629, so this does not look easy! In other words it is smaller than $\frac{1}{43}-\frac{1}{47}$ $\endgroup$
    – almagest
    Jan 26, 2020 at 20:52
  • $\begingroup$ You should check one of the (excellent) answers you have received for this question, for example that of Julian Rosen. It is a good habit, 3 weeks later, to check answers when you see you will not have better ones. I have seen that you have approximately been checking only, say, a half of the answers to your questions, some of them rather old... $\endgroup$
    – Jean Marie
    Feb 13, 2020 at 21:26

3 Answers 3

17
$\begingroup$

Because $1/x$ is a convex function, there is an inequality $1/x\leq \int_{x-1/2}^{x+1/2} t^{-1}\,dt$. It follows that $$ \frac{1}{18}+\frac{1}{19}+\ldots+\frac{1}{47}\leq \int_{17.5}^{47.5}\frac{dt}{t}=\ln\left(\frac{47.5}{17.5}\right). $$ By hand, we can check that $2.715\cdot 17.5=47.5125>47.5$, so that $\frac{47.5}{17.5}<2.715<e$, and therefore $$ \ln\left(\frac{47.5}{17.5}\right)<1. $$

$\endgroup$
2
  • 7
    $\begingroup$ By Euler's continued fraction for $e$, $$ e > 2 + \cfrac{1}{1 + \cfrac{1}{2 + \cfrac{1}{1 + \cfrac{1}{1}}}} = \frac{19}{7} = \frac{95}{35} = \frac{47.5}{17.5}. $$ $\endgroup$ Jan 26, 2020 at 23:36
  • $\begingroup$ Wow ! what an elegant answer. $\endgroup$
    – kaka
    Feb 7, 2020 at 11:07
6
$\begingroup$

The integer numbers $$n_k:=\biggl\lceil{10000\over k}\biggr\rceil\geq{10^4\over k}$$ are easy to calculate by hand for the thirty numbers $k\in[18\,..\,47]$. Since $$\sum_{k=18}^{47}\>n_k=9999<10^4$$ we arrive at the desired estimate.

$\endgroup$
6
$\begingroup$

In here, D.W. DeTemple established that the sequence

$$R_n = 1+\frac{1}{2}+\frac{1}{3} + \ldots + \frac{1}{n}-\ln\left(n+\frac{1}{2}\right)$$

is strictly decreasing and converges very fast toward the Euler-Mascheroni constant $\gamma$. Notice that the harmonic series

$$H_n = R_n + \ln\left(n+\frac{1}{2}\right)$$

So

$$R_{47} < R_{17} \Rightarrow H_{47}-\ln\left(47+\frac{1}{2}\right) < H_{17}-\ln\left(17+\frac{1}{2}\right)$$

Therefore

$$H_{47} - H_{17} < \ln\left(\frac{95}{2}\right)-\ln\left(\frac{35}{2}\right) < 1$$

The last inequality is equivalent with $95 < 35e$, which is true and can be verified by hand if we approximate $e$ to three decimals.

Notice that the same approach does not work with the classic definition

$$\gamma = \lim_{n \to \infty} \left(1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n}-\ln n\right)$$

because $H_n - \ln n$ converges very slow toward $\gamma$.

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .