Upper Bound For Sum Of Reciprocals: A Simpler Approach

Hey guys! Today, we're diving deep into a fascinating problem involving sequences, series, approximations, and finding upper bounds. It revolves around this intriguing sum: $\displaystyle \sum _{ \quad k \le a_n \\ \gcd(k,6)=1} \frac{1}{k}$. Basically, we want to figure out how big this sum can get, but in a nice, clean way. Let's break it down and make it super understandable.

Understanding the Sequence $a_n$

First, let's talk about the sequence $a_n$. It's defined as $a_n = 4n + 1 - 2 \displaystyle \left \lfloor \frac{n}{2} \right \rfloor$ for all natural numbers $n$ (that's just the integers greater than or equal to 1). What does this sequence actually do? Well, it generates integers that are odd and not divisible by 3. Think of it as a special number-generating machine! The first few terms of the sequence look like this: $a_1 = 3$, $a_2 = 5$, $a_3 = 7$, $a_4 = 9$, $a_5 = 11$, and so on. Notice anything? Yep, they're all odd numbers, and none of them can be divided evenly by 3. This sequence is crucial because it defines the upper limit of our summation. In essence, $a_n$ gives us a way to select which numbers we're going to add up the reciprocals of. Specifically, we're only interested in numbers less than or equal to $a_n$ that also satisfy a certain condition which we will explore next.

The Condition: $\gcd(k, 6) = 1$

Now, let's decode the condition $\gcd(k, 6) = 1$. The abbreviation "gcd" stands for greatest common divisor. So, this condition is saying that the greatest common divisor of $k$ and 6 must be 1. In simpler terms, $k$ and 6 should have no common factors other than 1. What does this mean for the numbers we're summing? Well, 6 has prime factors 2 and 3. So, $\gcd(k, 6) = 1$ means that $k$ cannot be divisible by 2 or 3. Aha! This confirms what we observed about the sequence $a_n$: its elements are odd and not divisible by 3. When we combine this gcd condition with the sequence $a_n$, we're essentially summing the reciprocals of all positive integers less than or equal to $a_n$ that are not divisible by 2 or 3. This is a pretty selective process, filtering out a lot of numbers.

The Summation: $\displaystyle \sum _{ \quad k \le a_n \\ \gcd(k,6)=1} \frac{1}{k}$

Okay, we've dissected the sequence $a_n$ and the gcd condition. Now, let's tackle the summation itself: $\displaystyle \sum _{ \quad k \le a_n \\ \gcd(k,6)=1} \frac{1}{k}$. This notation might look intimidating, but it's really just a fancy way of saying: "Add up the reciprocals (1/k) of all numbers k that meet our two conditions: k must be less than or equal to $a_n$, and the greatest common divisor of k and 6 must be 1." For instance, if we were looking at $a_5 = 11$, we would consider the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, and 11. We'd then filter out any numbers divisible by 2 or 3. This leaves us with 1, 5, 7, and 11. So, the sum would be $\frac{1}{1} + \frac{1}{5} + \frac{1}{7} + \frac{1}{11}$. Our main goal is to find a simpler upper bound for this sum. We don't need the exact value of the sum. We just need to find a number that we know the sum will never exceed. This is incredibly useful in many areas of math, especially when dealing with infinite series where calculating the exact sum might be impossible. Finding a good upper bound can tell us whether the series converges (approaches a finite value) or diverges (goes to infinity).

Finding a Simpler Upper Bound

Now for the fun part: finding a simpler upper bound! This involves a bit of mathematical trickery and clever estimation. The key idea here is to relate our somewhat complicated sum to something we already understand well, like the harmonic series or some variation thereof. We know that $\gcd(k, 6) = 1$ implies that $k$ is not divisible by 2 or 3. We can express the sum as:

$\displaystyle \sum _{ \quad k \le a_n \\ \gcd(k,6)=1} \frac{1}{k} = \sum_{k \le a_n} \frac{1}{k} - \sum_{k \le a_n, 2|k} \frac{1}{k} - \sum_{k \le a_n, 3|k} \frac{1}{k} + \sum_{k \le a_n, 6|k} \frac{1}{k}$

Here, we are including all numbers up to $a_n$, then subtracting the multiples of 2 and 3. However, we've subtracted the multiples of 6 twice (once as multiples of 2 and once as multiples of 3), so we need to add them back in. This is a classic application of the inclusion-exclusion principle. We can rewrite this as:

$\displaystyle \sum _{ \quad k \le a_n \\ \gcd(k,6)=1} \frac{1}{k} = \sum_{k \le a_n} \frac{1}{k} - \frac{1}{2}\sum_{k \le a_n/2} \frac{1}{k} - \frac{1}{3}\sum_{k \le a_n/3} \frac{1}{k} + \frac{1}{6}\sum_{k \le a_n/6} \frac{1}{k}$

We know that the harmonic sum $\sum_{k \le x} \frac{1}{k}$ can be approximated by $\ln(x) + \gamma + O(\frac{1}{x})$, where $\gamma$ is the Euler-Mascheroni constant (approximately 0.577). Using this approximation, we get:

$\displaystyle \sum _{ \quad k \le a_n \\ \gcd(k,6)=1} \frac{1}{k} \approx \ln(a_n) + \gamma - \frac{1}{2}(\ln(\frac{a_n}{2}) + \gamma) - \frac{1}{3}(\ln(\frac{a_n}{3}) + \gamma) + \frac{1}{6}(\ln(\frac{a_n}{6}) + \gamma)$

Simplifying this expression, we obtain:

$\displaystyle \sum _{ \quad k \le a_n \\ \gcd(k,6)=1} \frac{1}{k} \approx \ln(a_n) - \frac{1}{2}\ln(\frac{a_n}{2}) - \frac{1}{3}\ln(\frac{a_n}{3}) + \frac{1}{6}\ln(\frac{a_n}{6}) + \gamma(1 - \frac{1}{2} - \frac{1}{3} + \frac{1}{6})$

$\displaystyle \sum _{ \quad k \le a_n \\ \gcd(k,6)=1} \frac{1}{k} \approx \ln(a_n) - \frac{1}{2}\ln(a_n) + \frac{1}{2}\ln(2) - \frac{1}{3}\ln(a_n) + \frac{1}{3}\ln(3) + \frac{1}{6}\ln(a_n) - \frac{1}{6}\ln(6) + \frac{\gamma}{6}$

$\displaystyle \sum _{ \quad k \le a_n \\ \gcd(k,6)=1} \frac{1}{k} \approx (1 - \frac{1}{2} - \frac{1}{3} + \frac{1}{6})\ln(a_n) + \frac{1}{2}\ln(2) + \frac{1}{3}\ln(3) - \frac{1}{6}\ln(6) + \frac{\gamma}{6}$

$\displaystyle \sum _{ \quad k \le a_n \\ \gcd(k,6)=1} \frac{1}{k} \approx \frac{1}{6}\ln(a_n) + \frac{1}{2}\ln(2) + \frac{1}{3}\ln(3) - \frac{1}{6}(\ln(2) + \ln(3)) + \frac{\gamma}{6}$

$\displaystyle \sum _{ \quad k \le a_n \\ \gcd(k,6)=1} \frac{1}{k} \approx \frac{1}{6}\ln(a_n) + \frac{1}{3}\ln(2) + \frac{1}{6}\ln(3) + \frac{\gamma}{6}$

Since $a_n = 4n + 1 - 2 \lfloor \frac{n}{2} \rfloor$, we know that $2n - 1 \le a_n \le 4n + 1$. Thus, we can write:

$\displaystyle \sum _{ \quad k \le a_n \\ \gcd(k,6)=1} \frac{1}{k} \le \frac{1}{6}\ln(4n+1) + \frac{1}{3}\ln(2) + \frac{1}{6}\ln(3) + \frac{\gamma}{6}$

Therefore, a simpler upper bound can be expressed as:

$\frac{1}{6}\ln(4n+1) + C$, where $C = \frac{1}{3}\ln(2) + \frac{1}{6}\ln(3) + \frac{\gamma}{6} \approx 0.35$

In Conclusion: We have found a relatively simple upper bound for the sum $\displaystyle \sum _{ \quad k \le a_n \\ \gcd(k,6)=1} \frac{1}{k}$, which is approximately $\frac{1}{6}\ln(4n+1) + 0.35$. This bound provides a useful estimate of how the sum behaves as $n$ increases. Isn't that neat? I hope this breakdown helps you understand the problem better! Let me know if you have any questions.