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 '20 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 '20 at 21:26
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.

$\endgroup$
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 $e$, $$ e > 2 + \cfrac{1}{1 + \cfrac{1}{2 + \cfrac{1}{1 + \cfrac{1}{1}}}} = \frac{19}{7} = \frac{95}{35} = \frac{47.5}{17.5}. $$ $\endgroup$ – Calum Gilhooley Jan 26 '20 at 23:36
  • $\begingroup$ Wow ! what an elegant answer. $\endgroup$ – kaka Feb 7 '20 at 11:07
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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.