Question

Find a formula for the nth term of the sequence of partial sums \(\left\\{S_{n}\right\\} .\) Then evaluate lim \(S_{n}\) to obtain the value of the series or state that the series diverges.\(^{n \rightarrow \infty}\) $$\sum_{k=1}^{\infty}\left(\tan ^{-1}(k+1)-\tan ^{-1} k\right)$$

Short answer

Question: Determine whether the given series converges or diverges, and if it converges, find its value. $$\sum_{k=1}^{\infty}\left(\tan ^{-1}(k+1)-\tan ^{-1} k\right)$$ Answer: The series converges, and its value is $$\sum_{k=1}^{\infty}\left(\tan ^{-1}(k+1)-\tan ^{-1} k\right) = \frac{\pi}{2} - \tan^{-1}(1)$$

Step-by-step solution

Define the sequence of partial sums

In order to analyze this series, let's start by defining a sequence of partial sums, \(S_n\). A partial sum is simply the sum of the first n terms of the sequence. In this case, \(S_n\) is defined as: $$S_n = \sum_{k=1}^{n}\left(\tan ^{-1}(k+1)-\tan ^{-1} k\right)$$

Express the sequence of partial sums as a simplified formula

We will now look for a pattern that can help us simplify the expression for \(S_n\). Let's write out the first few terms of the series: $$\begin{aligned} S_n &= \left(\tan ^{-1}(2)-\tan ^{-1}(1)\right)+\left(\tan ^{-1}(3)-\tan ^{-1}(2)\right)+\left(\tan ^{-1}(4)-\tan ^{-1}(3)\right)+\cdots+\left(\tan ^{-1}(n+1)-\tan ^{-1}(n)\right) \end{aligned}$$ Notice how almost every term cancels with the previous term: $$S_n = \tan^{-1}(n+1) - \tan^{-1}(1)$$

Consider the limit as n approaches infinity

Now, we will consider the limit of \(S_n\) as n approaches infinity to determine if the series converges or diverges: $$\lim_{n \rightarrow \infty} S_n = \lim_{n \rightarrow \infty} \left( \tan^{-1}(n+1) - \tan^{-1}(1) \right)$$ Analyzing the individual terms as n approaches infinity, we have: $$\lim_{n \rightarrow \infty} \tan^{-1}(n+1) = \frac{\pi}{2} \text{ and } \lim_{n \rightarrow \infty} \tan^{-1}(1) = \tan^{-1}(1)$$ Since both limits exist, we can apply limit properties: $$\lim_{n \rightarrow \infty} S_n = \frac{\pi}{2} - \tan^{-1}(1)$$

Conclusion

Since the limit of the sequence of partial sums exists as n approaches infinity, the series converges. The value of the series is: $$\sum_{k=1}^{\infty}\left(\tan ^{-1}(k+1)-\tan ^{-1} k\right) = \frac{\pi}{2} - \tan^{-1}(1)$$

Key concepts

Partial Sums

Partial sums help us make sense of an infinite series by breaking it down into manageable parts. An infinite series is simply the sum of an infinite sequence of numbers. When we talk about partial sums, we take the first few terms of this sequence and add them together. This is called a partial sum. For example, if we have a series \( S_n = a_1 + a_2 + a_3 + \ldots + a_n \), the partial sum \( S_n \) is the sum of these first \( n \) terms.

Partial sums act as building blocks. They help us understand how an entire series behaves. By analyzing the sequence of partial sums, we can tell if a series converges or not. If the sequence of partial sums approaches a specific number as \( n \) becomes very large, then the series is said to converge, meaning it has a finite sum.

In the exercise, we had \( S_n = \sum_{k=1}^{n} \left(\tan^{-1}(k+1) - \tan^{-1}(k)\right) \). Here, each partial sum simplifies to \( S_n = \tan^{-1}(n+1) - \tan^{-1}(1) \) by canceling terms. This method of finding a pattern within partial sums is a crucial step in analyzing series behavior.

Convergence of Series

Convergence is a key concept when dealing with series. It tells us whether an infinite series has a finite sum. To determine convergence, we look at the sequence of partial sums. If the sequence of partial sums approaches a fixed number, we say the series converges. Otherwise, if it grows without bound, the series diverges.

When considering the convergence, the limit of the partial sums as \( n \to \infty \) is crucial. In our exercise, we looked at the limit \( \lim_{n \to \infty} S_n = \frac{\pi}{2} - \tan^{-1}(1) \), which exists. This means, as \( n \) gets larger, \( S_n \) approaches \( \frac{\pi}{2} - \tan^{-1}(1) \), confirming the series converges to this value.

Understanding convergence helps in many practical applications, such as calculating probabilities or solving differential equations, where infinite series often appear.

Telescoping Series

A telescoping series is a special type of series where many terms cancel out. This makes finding the sum easier. In a telescoping series, subsisting pairs of terms in the sequence may be designed to cancel each other out.

Let's consider our example: \( \sum_{k=1}^{\infty}\left(\tan^{-1}(k+1)-\tan^{-1} k\right) \). Each term can cancel with part of an adjacent term. Writing this out, you see: \[ \tan^{-1}(2) - \tan^{-1}(1), \tan^{-1}(3) - \tan^{-1}(2), \ldots, \tan^{-1}(n+1) - \tan^{-1}(n) \], the first \( \tan^{-1}(2) \) cancels with the \( \tan^{-1}(2) \) in the next pair, and so on.

Eventually, only the first part of the first term and the second part of the last term do not cancel. That's why telescoping reduces the problem to \( S_n = \tan^{-1}(n+1) - \tan^{-1}(1) \). This pattern is easy to spot and simplifies solving. It's an amazing trick that makes handling infinite series more approachable. By focusing on what remains after cancellation, we found the limit and showed our series converges neatly.