Proving Convergence Of The Sequence $a_{n+1} = A_n + A_n^2/n^2$

Hey guys! Ever stumbled upon a sequence that looks a bit intimidating but you just know there's a beautiful convergence hiding beneath the surface? Well, today we're diving deep into one such sequence. We're going to explore the ins and outs of proving its convergence, and trust me, it's a journey worth taking. Let's unravel the mysteries of the sequence defined by $a_1 = c ext{ where } c ext{ belongs to the interval } (0,1)$ and a_{n+1} = a_n + rac{a_n^2}{n^2} for all $n ext{ greater than or equal to } 1$. Our main goal? To show that the $\lim_{n \to \infty} a_n$ actually exists and is a nice, finite number. Buckle up, because we're about to get our math on!

Decoding the Sequence The Monotonicity Factor

When we're faced with a sequence, one of the first things we want to figure out is its behavior. Is it constantly increasing? Is it constantly decreasing? Or is it bouncing all over the place? In mathematical terms, we're investigating its monotonicity. For our sequence a_{n+1} = a_n + rac{a_n^2}{n^2}, let's break down why it's monotonically increasing.

The key here is the term $\frac{a_n^2}{n^2}$. Notice that $a_n^2$ is always non-negative (since anything squared is non-negative), and $n^2$ is also always positive (since n is a positive integer). This means that the entire fraction $\frac{a_n^2}{n^2}$ is non-negative. We're adding a non-negative quantity to $a_n$ to get $a_{n+1}$. Therefore, $a_{n+1}$ will always be greater than or equal to $a_n$. Mathematically, we can express this as $a_{n+1} \geq a_n$ for all $n \geq 1$. This, my friends, is the definition of a monotonically increasing sequence.

But wait, there's more! Since we know that $a_1 = c$ and $c$ is in the interval (0, 1), we know that $a_1$ is a positive number. And because we're always adding a non-negative term to get the next term in the sequence, all the terms in the sequence will be positive. This is an important piece of the puzzle that will help us later on.

So, to recap, we've established that our sequence is monotonically increasing and that all its terms are positive. This gives us a solid foundation for our next step proving the sequence is bounded.

Bounded Bliss Why Our Sequence Can't Run Wild

Okay, so we know our sequence is climbing upwards, but is it going to climb forever, shooting off into infinity? Or will it eventually hit a ceiling and settle down? This is where the concept of boundedness comes in. A sequence is bounded if its terms are all less than some fixed number (an upper bound) and greater than some fixed number (a lower bound). We already know our sequence is bounded below by 0, since all its terms are positive. The real challenge is to show that it's bounded above.

To tackle this, we're going to use a clever trick involving an auxiliary sequence. Think of it as a mathematical sidekick, helping us out with the main problem. The auxiliary sequence we'll use is $b_n = \frac{1}{a_n}$. Why this one? Well, stick around and you'll see!

Let's take the reciprocal of our original recurrence relation:

$\frac{1}{a_{n+1}} = \frac{1}{a_n + \frac{a_n^2}{n^2}}$

Now, let's do a little algebraic dance to massage this equation into a more helpful form. We can rewrite the right-hand side by finding a common denominator:

$\frac{1}{a_{n+1}} = \frac{1}{\frac{n^2 a_n + a_n^2}{n^2}} = \frac{n^2}{n^2 a_n + a_n^2}$

Taking the reciprocal again, we get:

$b_{n+1} = \frac{1}{a_{n+1}} = \frac{n^2}{a_n(n^2 + a_n)}$

Now, let's look at the difference between consecutive terms of our auxiliary sequence:

$b_{n+1} - b_n = \frac{1}{a_{n+1}} - \frac{1}{a_n} = \frac{n^2}{n^2 a_n + a_n^2} - \frac{1}{a_n}$

Combining the fractions, we get:

$b_{n+1} - b_n = \frac{n^2 - (n^2 + a_n)}{a_n(n^2 + a_n)} = \frac{-a_n}{a_n(n^2 + a_n)} = \frac{-1}{n^2 + a_n}$

Since $n^2$ and $a_n$ are both positive, the denominator $n^2 + a_n$ is positive. This means that $b_{n+1} - b_n$ is negative. In other words, $b_{n+1} < b_n$, so the auxiliary sequence $b_n$ is monotonically decreasing. Awesome! This is a crucial insight.

Next, we'll use this information to find an inequality that helps us bound $b_n$.

Since $b_{k+1} - b_k = \frac{-1}{k^2 + a_k}$, we can say:

$b_{k+1} - b_k < \frac{-1}{k^2 + 0} = -\frac{1}{k^2}$ (Because $a_k$ is positive, $k^2 + a_k > k^2$)

Now, let's sum up these differences from k = 1 to n-1:

$\sum_{k=1}^{n-1} (b_{k+1} - b_k) < \sum_{k=1}^{n-1} -\frac{1}{k^2}$

The left-hand side telescopes (meaning most of the terms cancel out!), leaving us with:

$b_n - b_1 < - \sum_{k=1}^{n-1} \frac{1}{k^2}$

Multiplying both sides by -1 and flipping the inequality, we get:

$b_1 - b_n > \sum_{k=1}^{n-1} \frac{1}{k^2}$

Adding $b_n$ to both sides and rearranging, we get:

$b_1 > b_n + \sum_{k=1}^{n-1} \frac{1}{k^2}$

Now, let's add $\frac{1}{n^2}$ to the summation and subtract it from the left side:

$b_1 - \frac{1}{n^2} > b_n + \sum_{k=1}^{n} \frac{1}{k^2} - \frac{1}{n^2}$

Rearranging, we get:

$b_n < b_1 - \sum_{k=1}^{n} \frac{1}{k^2} + \frac{1}{n^2}$

Since $\sum_{k=1}^{\infty} \frac{1}{k^2}$ converges (it's a well-known p-series with p = 2), it's bounded by some constant, let's call it $S$. Also, $\frac{1}{n^2}$ goes to 0 as n goes to infinity.

Therefore, there is some number $M$ such that $b_n$ is greater than $M$. That is $M < b_n$. Taking reciprocal gives us the upper bound for $a_n$ as $a_n < \frac{1}{M}$.

But this means there's a limit! As $n$ approaches infinity, $a_n$ will get closer and closer to this upper bound, but it will never surpass it. This, my friends, is the essence of convergence.

Convergence Conquered The Grand Finale

We've reached the climax of our mathematical adventure! We've shown that our sequence $a_{n+1} = a_n + \frac{a_n^2}{n^2}$ is both monotonically increasing and bounded above. And what does this tell us? Drumroll please...

The Monotone Convergence Theorem states that if a sequence is monotonically increasing and bounded above, then it converges to a finite limit. Bam! We've proven that $\lim_{n \to \infty} a_n$ exists and is finite.

We navigated through the intricacies of monotonicity, employed a clever auxiliary sequence to establish boundedness, and finally, triumphantly invoked the Monotone Convergence Theorem to seal the deal. High fives all around!

So, the next time you encounter a seemingly complex sequence, remember the tools we've used today. Break it down, analyze its behavior, and don't be afraid to bring in a mathematical sidekick or two. Convergence might just be around the corner!

Repair Input Keyword

Prove that the sequence defined by $a_1 = c \in (0,1)$ and $a_{n+1} = a_n + \frac{a_n^2}{n^2}$ for all $n \geq 1$ converges and find its finite limit.