Is The Sequence Decreasing? A Proof

Hey guys, welcome back to Plastik Magazine! Today, we're diving deep into the fascinating world of sequences and series, a topic that might sound a bit intimidating at first, but trust me, it's super cool once you get the hang of it. We're going to tackle a specific recursive sequence and figure out if it's decreasing. So, grab your notebooks (or just relax and read along!), because we're about to break down this mathematical puzzle step-by-step. Our main focus today is to rigorously prove whether the sequence defined by $a_1 = 1$ and $a_k = \sum\limits_{i=1}^{k-1}a_i\left(\frac{1}{k-i}-\frac{1}{k-i+1}\right)$ for $k \ge 2$ is a decreasing sequence. This means we need to show that each term is less than or equal to the previous term, i.e., $a_k \le a_{k-1}$ for all $k \ge 2$. This is a common problem in discrete mathematics, and understanding how to prove such properties is crucial for a lot of advanced mathematical concepts. We'll be exploring the relationship between terms and using the given recursive definition to our advantage. So, let's get started and unravel this mathematical mystery together!

Understanding the Recursive Definition

Alright, let's get down to business and really understand what this recursive definition is telling us. We're given $a_1 = 1$, which is our starting point. Then, for any term $a_k$ where $k$ is greater than or equal to 2, its value is determined by a sum involving all the previous terms ($a_1$ through $a_{k-1}$). This is the essence of a recursive definition – each term depends on the ones that came before it. The specific formula for $a_k$ is $a_k = \sum\limits_{i=1}^{k-1}a_i\left(\frac{1}{k-i}-\frac{1}{k-i+1}\right)$. Let's unpack that summation part. The term inside the sum has $a_i$ multiplied by a difference of two fractions: $\left(\frac{1}{k-i}-\frac{1}{k-i+1}\right)$. This difference of fractions is quite interesting. If we find a common denominator, which is $(k-i)(k-i+1)$, the expression becomes $\frac{(k-i+1)-(k-i)}{(k-i)(k-i+1)} = \frac{1}{(k-i)(k-i+1)}$. So, the formula for $a_k$ can be rewritten as $a_k = \sum\limits_{i=1}^{k-1}\frac{a_i}{(k-i)(k-i+1)}$. This simplified form might make it easier to work with. It tells us that each new term $a_k$ is a weighted sum of all preceding terms, where the weights are given by $\frac{1}{(k-i)(k-i+1)}$. Notice that for a fixed $k$, as $i$ increases, the denominator $(k-i)(k-i+1)$ changes. Let's look at the first few terms to get a feel for this sequence. We know $a_1 = 1$. For $k=2$, we have $a_2 = \sum\limits_{i=1}^{1}a_i\left(\frac{1}{2-i}-\frac{1}{2-i+1}\right) = a_1\left(\frac{1}{2-1}-\frac{1}{2-1+1}\right) = a_1\left(\frac{1}{1}-\frac{1}{2}\right) = 1\left(\frac{1}{2}\right) = \frac{1}{2}$. So, $a_2 = \frac{1}{2}$. Now for $k=3$, $a_3 = \sum\limits_{i=1}^{2}a_i\left(\frac{1}{3-i}-\frac{1}{3-i+1}\right)$. This expands to $a_1\left(\frac{1}{3-1}-\frac{1}{3-1+1}\right) + a_2\left(\frac{1}{3-2}-\frac{1}{3-2+1}\right) = a_1\left(\frac{1}{2}-\frac{1}{3}\right) + a_2\left(\frac{1}{1}-\frac{1}{2}\right)$. Substituting the values we know: $a_3 = 1\left(\frac{1}{6}\right) + \frac{1}{2}\left(\frac{1}{2}\right) = \frac{1}{6} + \frac{1}{4} = \frac{2+3}{12} = \frac{5}{12}$. So, $a_3 = \frac{5}{12}$. Let's compare these terms: $a_1 = 1$, $a_2 = \frac{1}{2}$, $a_3 = \frac{5}{12}$. Since $1 > \frac{1}{2}$ and $\frac{1}{2} = \frac{6}{12} > \frac{5}{12}$, it looks like the sequence is decreasing. But we need a formal proof that holds for all $k$. The structure of the sum, involving differences of reciprocals, is a common pattern in calculus and discrete math, often related to telescoping sums or approximations. Here, the term $\frac{1}{k-i}-\frac{1}{k-i+1}$ can be seen as a small quantity that decreases as $k-i$ increases. This hints that later terms in the sum might contribute less, or that the weights themselves decrease in a specific way. Understanding these nuances is key to proving the decreasing nature of the sequence.

Proving the Sequence is Decreasing: The Strategy

Now, let's talk strategy for proving that our sequence $a_k$} is indeed decreasing. To prove that a sequence {$a_k$} is decreasing, we need to show that $a_k \le a_{k-1}$ for all $k \ge 2$. This is our ultimate goal. We've already calculated the first few terms ($a_1=1$, $a_2=1/2$, $a_3=5/12$) and they seem to follow this pattern. However, simply looking at the first few terms isn't enough for a mathematical proof. We need a method that works for any $k$. There are a few common approaches to proving properties of recursively defined sequences. One powerful technique is mathematical induction. This involves proving a base case (which we've already started with $k=2$) and then assuming the property holds for some arbitrary integer $m$ (the inductive hypothesis) and proving it also holds for $m+1$. Another approach is to directly manipulate the recursive definition. We can write out the definitions for $a_k$ and $a_{k-1}$ and try to show that the difference $a_k - a_{k-1}$ is less than or equal to zero. Let's consider the direct manipulation approach first, as it often gives more insight into the structure of the sequence. We have the definition for $a_k$ $a_k = \sum\limits_{i=1^k-1}a_i\left(\frac{1}{k-i}-\frac{1}{k-i+1}\right)$. And for $a_{k-1}$, we would replace $k$ with $k-1$ $a_{k-1 = \sum\limits_i=1}^{k-2}a_i\left(\frac{1}{(k-1)-i}-\frac{1}{(k-1)-i+1}\right) = \sum\limits_{i=1}^{k-2}a_i\left(\frac{1}{k-1-i}-\frac{1}{k-i}\right)$. This looks a bit complicated to compare directly. The indices and the difference terms don't line up nicely. Let's go back to the simplified form of $a_k$ $a_k = \sum\limits_{i=1^{k-1}\frac{a_i}{(k-i)(k-i+1)}$. This might be more manageable. The challenge here is that $a_i$ itself is defined recursively. So, proving a_k le a_{k-1} directly might involve a lot of algebraic manipulation. A key insight might come from examining the terms $a_i$ inside the sum. If we can establish some bounds or properties for $a_i$ itself, it could simplify the proof. For example, if we can show that $a_i > 0$ for all $i$, and that the weights $\frac{1}{(k-i)(k-i+1)}$ decrease in a certain way as $i$ increases for a fixed $k$, it might lead us to the solution. The expression $\frac{1}{(k-i)(k-i+1)}$ is always positive for $i < k$. So, $a_k$ is a sum of positive terms (assuming $a_i > 0$). This suggests that if $a_1 > 0$, then all subsequent terms will also be positive. This is a good starting point! Now, let's think about how to use induction. For induction, we'd first show a_2 le a_1. We have $a_1=1$ and $a_2=1/2$, so 1/2 le 1, which is true. The inductive hypothesis would be to assume a_m le a_{m-1} for all 2 le m le k. Then we'd need to prove a_{k+1} le a_k. This requires substituting the full recursive definition into the inequality, which can become quite dense. A more promising approach might be to try and find a simpler form or relation for $a_k$. Sometimes, rewriting the original recurrence can lead to a breakthrough. Let's focus on the structure of the sum again. The term $\left(\frac{1}{k-i}-\frac{1}{k-i+1}\right)$ is a difference of consecutive terms in the harmonic sequence. This structure is often found in problems that simplify nicely.

Simplifying the Recursive Relation

Let's try to simplify the recursive relation to make the proof more straightforward. We have $a_k = \sum\limits_{i=1}^{k-1}a_i\left(\frac{1}{k-i}-\frac{1}{k-i+1}\right)$. As we saw, the term in the parenthesis simplifies to $\frac{1}{(k-i)(k-i+1)}$. So, $a_k = \sum\limits_{i=1}^{k-1}\frac{a_i}{(k-i)(k-i+1)}$. Now, let's consider the term $a_k$ for $k+1$: $a_{k+1} = \sum\limits_{i=1}^{k}a_i\left(\frac{1}{(k+1)-i}-\frac{1}{(k+1)-i+1}\right) = \sum\limits_{i=1}^{k}\frac{a_i}{(k+1-i)(k+2-i)}$. We want to show $a_{k+1} \le a_k$. This still looks challenging. What if we try to rearrange the original equation differently? Let's look at the terms in the sum for $a_k$: $a_k = a_1\left(\frac{1}{k-1}-\frac{1}{k}\right) + a_2\left(\frac{1}{k-2}-\frac{1}{k-1}\right) + \dots + a_{k-1}\left(\frac{1}{1}-\frac{1}{2}\right)$. This is $a_k = a_1\left(\frac{1}{(k-1)k}\right) + a_2\left(\frac{1}{(k-2)(k-1)}\right) + \dots + a_{k-1}\left(\frac{1}{1 cdot 2}\right)$. Let's try to find a relation between $a_k$ and $a_{k-1}$ by manipulating the definition. Consider the expression for $a_k$ and $a_{k-1}$. a_k = rac{a_1}{k-1} - rac{a_1}{k} + rac{a_2}{k-2} - rac{a_2}{k-1} + rac{a_3}{k-3} - rac{a_3}{k-2} + inom{ ext{general term } rac{a_i}{k-i} - rac{a_i}{k-i+1}}{ ext{ and so on}} + rac{a_{k-1}}{1} - rac{a_{k-1}}{2}. This expansion doesn't immediately reveal a telescoping sum. However, let's consider a different perspective. What if we look at the difference $a_k - a_{k-1}$? This seems to be the most direct approach. We have a_k = rac{1}{2}a_1 + rac{1}{6}a_2 + rac{1}{12}a_3 + inom{ ext{general term } rac{a_i}{(k-i)(k-i+1)}}{ ext{ and so on}}. This is getting messy. Let's try to find a different way to express $a_k$. Sometimes, these recursive sequences have a hidden closed-form or a simpler recurrence. Let's consider the quantity k rac{a_k}{a_{k-1}} or similar ratios. But we want to prove a_k le a_{k-1}. Let's rewrite the sum for $a_k$ in a way that groups terms differently. a_k = rac{a_1}{k(k-1)} + rac{a_2}{(k-1)(k-2)} + rac{a_3}{(k-2)(k-3)} + inom{ ext{general term } rac{a_{k-j}}{j(j+1)} ext{ where } j=k-i}}{ ext{ and so on}}. Let's rewrite the definition by splitting the sum: a_k = rac{a_1}{k-1} - rac{a_1}{k} + rac{a_2}{k-2} - rac{a_2}{k-1} + rac{a_3}{k-3} - rac{a_3}{k-2} + inom{ ext{general term } rac{a_i}{k-i} - rac{a_i}{k-i+1}}{ ext{ and so on}}. Let's rearrange the terms by grouping $a_i$ terms: a_k = -rac{a_1}{k} + rac{a_1 - a_2}{k-1} + rac{a_2 - a_3}{k-2} + rac{a_3 - a_4}{k-3} + inom{ ext{general term } rac{a_{i}-a_{i+1}}{k-i}}{ ext{ and so on}} + rac{a_{k-1}}{1}. This is incorrect rearrangement. Let's use the simplified form a_k = rac{a_1}{(k-1)k} + rac{a_2}{(k-2)(k-1)} + rac{a_3}{(k-3)(k-2)} + inom{ ext{general term } rac{a_{k-j}}{j(j+1)} ext{ for } j=1, inom{ ext{to } k-1}}{ ext{ and so on}}. Let's try to express $a_k$ in terms of $a_{k-1}$ and earlier terms. a_k = rac{a_1}{(k-1)k} + rac{a_2}{(k-2)(k-1)} + inom{ ext{general term } rac{a_{k-2}}{(2)(3)}}{ ext{ and so on}} + rac{a_{k-1}}{2 imes 1}. This sum is structured in a way that might lead to cancellation if we look at differences. Let's try to prove a_k le a_{k-1} using induction. Base case: a_2 = 1/2 le a_1 = 1. True. Inductive Hypothesis (IH): Assume a_m le a_{m-1} for 2 le m le k. We need to show a_{k+1} le a_k. $a_{k+1} = \sum\limits_{i=1}^{k}\frac{a_i}{(k+1-i)(k+2-i)}$. This still seems complex. Let's try to find a relationship that is easier to prove. What if we try to find an expression for $a_k$ that doesn't involve a sum directly? Let's re-examine a_k = rac{a_1}{k-1} - rac{a_1}{k} + rac{a_2}{k-2} - rac{a_2}{k-1} + inom{ ext{general term } rac{a_i}{k-i} - rac{a_i}{k-i+1}}{ ext{ and so on}}. Let $j=k-i$. Then $i=k-j$. When $i=1$, $j=k-1$. When $i=k-1$, $j=1$. So $a_k = \sum\limits_{j=1}^{k-1} a_{k-j} \left(\frac{1}{j} - \frac{1}{j+1}\right) = \sum\limits_{j=1}^{k-1} \frac{a_{k-j}}{j(j+1)}$. This form looks more manageable for comparing terms. Let's use this form: a_k = rac{a_{k-1}}{1 imes 2} + rac{a_{k-2}}{2 imes 3} + rac{a_{k-3}}{3 imes 4} + inom{ ext{general term } rac{a_{k-j}}{j(j+1)}}{ ext{ and so on}} + rac{a_1}{(k-1)k}. Now let's write $a_{k+1}$ using this form: a_{k+1} = \sum\limits_{j=1}^{k} \frac{a_{k+1-j}}{j(j+1)} = rac{a_k}{1 imes 2} + rac{a_{k-1}}{2 imes 3} + rac{a_{k-2}}{3 imes 4} + inom{ ext{general term } rac{a_{k+1-j}}{j(j+1)}}{ ext{ and so on}} + rac{a_1}{k(k+1)}. We want to show a_{k+1} le a_k. Using the new form: a_{k+1} = rac{a_k}{2} + rac{a_{k-1}}{6} + rac{a_{k-2}}{12} + inom{ ext{terms with smaller } a_i}{ ext{ and so on}}. And a_k = rac{a_{k-1}}{2} + rac{a_{k-2}}{6} + inom{ ext{terms with smaller } a_i}{ ext{ and so on}}. This still isn't directly showing a_{k+1} le a_k. However, notice that $\frac{1}{j(j+1)}$ decreases as $j$ increases. This means the terms with larger $j$ (which correspond to earlier $a_i$'s) have smaller weights. This structure is key. Let's try to prove $a_k > 0$ for all $k$ by induction. Base case $a_1 = 1 > 0$. Assume $a_i > 0$ for 1 le i le k-1. Then $a_k = \sum\limits_{i=1}^{k-1}\frac{a_i}{(k-i)(k-i+1)}$. Since $a_i > 0$ and $\frac{1}{(k-i)(k-i+1)} > 0$, their product is positive, and the sum of positive terms is positive. Thus, $a_k > 0$ for all $k$. Now, let's analyze $a_k - a_{k-1}$. This direct approach is proving tricky. Let's consider the possibility that the relation simplifies further or there's a trick. What if we look at the sum \sum_{i=1}^{k-1} a_i rac{1}{(k-i)(k-i+1)}? The terms $\frac{1}{(k-i)(k-i+1)}$ form a decreasing sequence of weights for $a_i$ as $i$ increases from $1$ to $k-1$. That is, $a_1$ is multiplied by $\frac{1}{(k-1)k}$, $a_2$ by $\frac{1}{(k-2)(k-1)}$, ..., $a_{k-1}$ by $\frac{1}{1 imes 2}$. The coefficient for $a_{k-1}$ is the largest. This might be useful. The simplified form a_k = rac{a_{k-1}}{2} + rac{a_{k-2}}{6} + inom{ ext{terms with smaller } a_i}{ ext{ and so on}} is very suggestive. If $a_k$ were simply proportional to $a_{k-1}$ with a coefficient less than 1, it would be decreasing. Here, it's a sum. Let's try to use induction more rigorously on the inequality a_k le a_{k-1}.

The Proof by Induction

Okay guys, we've explored the definitions and tried a few angles. The most robust way to prove this property for all $k$ is usually mathematical induction. We've already established our base case: $a_1 = 1$ and $a_2 = 1/2$. Clearly, a_2 le a_1 since 1/2 le 1. So, our base case for $k=2$ holds. Now, for the inductive hypothesis, let's assume that the sequence is decreasing up to some integer m le k. That is, assume a_j le a_{j-1} for all 2 le j le m. Our goal is to prove that a_{m+1} le a_m. Let's use the definition: $a_{m+1} = \sum\limits_{i=1}^{m}a_i\left(\frac{1}{(m+1)-i}-\frac{1}{(m+1)-i+1}\right) = \sum\limits_{i=1}^{m}\frac{a_i}{((m+1)-i)((m+1)-i+1)}$. And $a_m = \sum\limits_{i=1}^{m-1}a_i\left(\frac{1}{m-i}-\frac{1}{m-i+1}\right) = \sum\limits_{i=1}^{m-1}\frac{a_i}{(m-i)(m-i+1)}$. Comparing these two sums directly is still a bit of a beast. However, we know from our inductive hypothesis that a_i le a_{i-1} for i le m. This means that the terms a_1, a_2, inom{ ext{up to }}{ ext{ and so on}} a_m are in non-increasing order. Let's consider the terms in the sum for $a_{m+1}$: a_{m+1} = rac{a_1}{(m)(m+1)} + rac{a_2}{(m-1)(m)} + rac{a_3}{(m-2)(m-1)} + inom{ ext{general term } rac{a_i}{(m+1-i)(m+2-i)}}{ ext{ and so on}} + rac{a_m}{1 imes 2}. Let's rewrite the sum for $a_m$ using the same denominator structure for clarity, though the upper limit is $m-1$: a_m = rac{a_1}{(m-1)m} + rac{a_2}{(m-2)(m-1)} + inom{ ext{general term } rac{a_i}{(m-i)(m-i+1)}}{ ext{ and so on}} + rac{a_{m-1}}{1 imes 2}. The challenge is relating the weights. Let's try a different inductive approach. Maybe we can prove a stronger property. Let's consider the quantity $a_k - a_{k-1}$ again. Using the form a_k = rac{a_{k-1}}{2} + rac{a_{k-2}}{6} + rac{a_{k-3}}{12} + inom{ ext{terms with smaller } a_i}{ ext{ and so on}}: a_k = rac{a_{k-1}}{1 imes 2} + rac{a_{k-2}}{2 imes 3} + inom{ ext{general term } rac{a_{k-j}}{j(j+1)} ext{ for } j=1, inom{ ext{to } k-1}}{ ext{ and so on}}. We want to show a_k le a_{k-1}. Consider the difference: a_k - a_{k-1} = rac{a_{k-1}}{2} + rac{a_{k-2}}{6} + inom{ ext{smaller terms}}{ ext{ and so on}} - a_{k-1} = -rac{1}{2}a_{k-1} + rac{a_{k-2}}{6} + inom{ ext{smaller terms}}{ ext{ and so on}}. This doesn't immediately show the difference is negative. However, let's reconsider the original form: a_k = \sum\limits_{i=1}^{k-1}a_iulletrac{1}{(k-i)(k-i+1)}. Let's try to show that $a_k$ is a weighted average of previous terms with weights that sum to less than 1, which would imply $a_k < ext{max}(a_i)$. But this doesn't directly prove a_k le a_{k-1}. Let's focus on the simplified sum again: a_k = rac{a_{k-1}}{1 imes 2} + rac{a_{k-2}}{2 imes 3} + inom{ ext{general term } rac{a_{k-j}}{j(j+1)} ext{ for } j=1, inom{ ext{to } k-1}}{ ext{ and so on}}. We want to prove a_k le a_{k-1}. Let's assume, for the sake of induction, that a_i le a_{i-1} for all i le k-1. This means a_{k-1} le a_{k-2} le inom{ ext{and so on}}{ ext{ and so on}} le a_1. Therefore, a_i le a_{k-1} for all i le k-1. Now, let's use this in the expression for $a_k$: a_k = rac{a_{k-1}}{1 imes 2} + rac{a_{k-2}}{2 imes 3} + rac{a_{k-3}}{3 imes 4} + inom{ ext{general term } rac{a_{k-j}}{j(j+1)}}{ ext{ and so on}} + rac{a_1}{(k-1)k}. Since a_i le a_{k-1} for all i le k-1, we can write: a_k le rac{a_{k-1}}{1 imes 2} + rac{a_{k-1}}{2 imes 3} + rac{a_{k-1}}{3 imes 4} + inom{ ext{general term } rac{a_{k-1}}{j(j+1)}}{ ext{ and so on}} + rac{a_{k-1}}{(k-1)k}. This simplifies to: a_k le a_{k-1} inom{\sum\limits_{j=1}^{k-1} \frac{1}{j(j+1)}} \$}. Now, the sum $\sum\limits_{j=1}^{k-1} \frac{1}{j(j+1)}$ is a telescoping sum! $\frac{1}{j(j+1)} = \frac{1}{j} - \frac{1}{j+1}$. So, \sum\limits_{j=1}^{k-1} \left(\frac{1}{j} - \frac{1}{j+1}\right) = \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + inom{ ext{and so on}}{ ext{ and so on}} + \left(\frac{1}{k-1} - \frac{1}{k}\right) = 1 - \frac{1}{k}. Therefore, a_k le a_{k-1} \left(1 - \frac{1}{k}\right). Since $1 - \frac{1}{k} < 1$ for k le 2, this implies $a_k < a_{k-1}$ for k le 2. However, this is not the final step. This inequality a_k le a_{k-1} (1 - rac{1}{k}) does show that $a_k$ is strictly less than $a_{k-1}$ for k le 2, assuming $a_{k-1} > 0$ (which we've proved). So, this means $a_k < a_{k-1}$ for k le 2. This directly proves that the sequence is strictly decreasing for k le 2. The inductive step holds. Thus, by induction, the sequence {$a_k$} is strictly decreasing.

Conclusion

So there you have it, guys! We've successfully proven that the sequence defined by $a_1 = 1$ and $a_k = \sum\limits_{i=1}^{k-1}a_i\left(\frac{1}{k-i}-\frac{1}{k-i+1}\right)$ for $k \ge 2$ is indeed a decreasing sequence. We achieved this by first simplifying the recursive formula and then employing a powerful proof technique: mathematical induction. We showed that $a_2 < a_1$, establishing the base case. Then, we used the inductive hypothesis (assuming the sequence is decreasing up to $a_m$) to prove that $a_{m+1} < a_m$. The key insight came from rewriting $a_k$ as $a_k = \sum\limits_{j=1}^{k-1} \frac{a_{k-j}}{j(j+1)}$ and using the inductive hypothesis a_i le a_{k-1} for i le k-1 to establish a_k le a_{k-1} \sum\limits_{j=1}^{k-1} \frac{1}{j(j+1)}. The telescoping sum of $\frac{1}{j(j+1)}$ simplified this to a_k le a_{k-1} (1 - rac{1}{k}), which proves the decreasing nature. This kind of rigorous proof is fundamental in discrete mathematics and helps us understand the behavior of sequences and functions. It's amazing how a seemingly complex formula can be analyzed and its properties unveiled with the right tools. Keep exploring, keep questioning, and I'll catch you in the next article!