Question

For the following telescoping series, 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

Answer: The given telescoping series converges, and its value is \(\frac{\pi}{4}\).

Step-by-step solution

Find the series term

The given infinite series can be represented as: $$\sum_{k=1}^{\infty} \left(\tan^{-1}(k+1) - \tan^{-1}(k)\right)$$

Representation as partial sums

To translate this representation as partial sums, we have: $$S_n = \sum_{k=1}^{n} \left(\tan^{-1}(k+1) - \tan^{-1}(k)\right)$$

Telescoping

Now we will telescope the series by canceling out common terms in consecutive elements of the sum: $$\begin{aligned} S_n &= \left(\tan^{-1}(2) - \tan^{-1}(1)\right) + \left(\tan^{-1}(3) - \tan^{-1}(2)\right) + \cdots + \left(\tan^{-1}(n+1) - \tan^{-1}(n)\right) \\ &= \left(\tan^{-1}(n+1) - \tan^{-1}(1)\right) \end{aligned}$$

Find the limit

Now, we want to find the value of the series by evaluating lim \(S_n\) as \(n\) approaches infinity: $$\lim_{n \to \infty} S_n = \lim_{n \to \infty} \left(\tan^{-1}(n+1) - \tan^{-1}(1)\right)$$ Notice that as n goes to infinity, \((n+1)\) grows towards infinity, and \(\tan^{-1}(n+1)\) approaches \(\frac{\pi}{2}\) since the range of the inverse tangent function is from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\). Thus, $$\lim_{n \to \infty} S_n = \lim_{n \to \infty} \left(\frac{\pi}{2} - \tan^{-1}(1)\right) = \frac{\pi}{2} - \tan^{-1}(1)$$ Since \(\tan^{-1}(1) = \frac{\pi}{4}\), we get: $$\lim_{n \to \infty} S_n = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$$ Therefore, the given series converges, and its value is \(\frac{\pi}{4}\).

Key concepts

Sequence of Partial Sums

Imagine a series as a long string of numbers added together, an infinite string in this case. When you're working with a telescoping series, like the one given in our exercise, the idea is to find a simpler way to express this long sum. This is where the sequence of partial sums comes into play.

You don't start by tackling the infinite series directly. Instead, you evaluate smaller parts of it, called partial sums. A partial sum, denoted by \(S_n\), is simply the sum of the first \(n\) terms of your series. For example, if you have a series \(a_1 + a_2 + a_3 + \ldots\), the partial sum \(S_3\) would be \(a_1 + a_2 + a_3\).

In the telescoping series given:

• We express the partial sum \(S_n\) using the series terms from the first term up to the \(n\)-th term.
• Consequently, for our exercise, \(S_n\) transforms from an infinite sum into a more manageable expression that mirrors the cancellations typical of a telescoping series.

This simplification allows us to see the broader picture of how the sums behave as \(n\) becomes very large.

Inverse Trigonometric Functions

The inverse trigonometric functions play a crucial role in many mathematical expressions, particularly when dealing with series like the one in our exercise. Specifically, we're working with the inverse tangent function, \(\tan^{-1}(x)\), also known as the arctangent function.

The inverse tangent function is the inverse operation of the tangent function, essentially solving for the angle whose tangent is a particular value. It has a unique impact on our series, given its predictable range of values:

• The range of \(\tan^{-1}(x)\) is from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\).
• This means that as \(x\) increases to positive infinity, the value of \(\tan^{-1}(x)\) approaches \(\frac{\pi}{2}\).

In the context of the telescoping series in this exercise, this property is key because it explains why \(\tan^{-1}(k+1) - \tan^{-1}(k)\) simplifies further when expressed as partial sums. Understanding this function's behavior is critical as it directly influences the limit and convergence of the series as \(n\) approaches infinity.

Series Convergence

Convergence is the term used to describe whether a series approaches a specific value as its terms stretch toward infinity. It answers the essential question: does the series settle down to a particular number?

In the case of our telescoping series, after simplifying the sequence of partial sums with the help of inverse trigonometric functions, the issue of convergence becomes apparent.

• We calculate the limit of the partial sum \(\lim_{n \to \infty} S_n\).
• The inverse tangent function's nature aids us, as it guarantees that \(\tan^{-1}(n+1)\) approaches \(\frac{\pi}{2}\) as \(n\) grows indefinitely.

This means the partial sum equation \(\tan^{-1}(n+1) - \tan^{-1}(1)\) ultimately resolves to a fixed number as \(n\) heads toward infinity. The series converges to \(\frac{\pi}{4}\), establishing that our original infinite series isn't merely an accumulation of infinite values—it actually locks onto a precise value, expressing that the sum is stable and predictable over the long run.