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.\(^{n \rightarrow \infty}\) $$\sum_{k=1}^{\infty}(\sqrt{k+1}-\sqrt{k})$$

Short answer

Answer: The series diverges.

Step-by-step solution

Find the nth partial sum \(S_n\)

To find the nth partial sum, take the sum of the first n terms of the sequence. $$S_n = \sum_{k=1}^{n}(\sqrt{k+1}-\sqrt{k})$$

Simplify the sum

Notice that \(\sqrt{k+1} - \sqrt{k}\) is a telescoping series since every term cancels out with the term before it. \begin{align*} S_n &= (\sqrt{2}-\sqrt{1})+(\sqrt{3}-\sqrt{2})+(\sqrt{4}-\sqrt{3})+\cdots+(\sqrt{n+1}-\sqrt{n}) \\ S_n &= -\sqrt{1}+\sqrt{2} -\sqrt{2}+\sqrt{3}+\cdots-\sqrt{n}+\sqrt{n+1} \\ S_n &= \sqrt{n+1} - \sqrt{1} \end{align*}

Find the limit of \(S_n\) as \(n \rightarrow \infty\)

To find the limit of \(S_n\) as \(n\rightarrow\infty\), consider the formula we derived for \(S_n\). $$\lim_{n\rightarrow\infty} S_n = \lim_{n\rightarrow\infty}(\sqrt{n+1} - \sqrt{1})$$ Using properties of limits, we can rewrite as: \begin{align*} \lim_{n\rightarrow\infty}(\sqrt{n+1} - \sqrt{1}) &= \lim_{n\rightarrow\infty}\sqrt{n+1} - \lim_{n\rightarrow\infty}\sqrt{1} \\ &= \infty - 1 \\ &= \infty \end{align*}

Determine the convergence or divergence of the series

Since \(\lim_{n\rightarrow\infty}S_n = \infty\), it shows that the series does not converge to a finite value. Therefore, we can conclude that the series diverges.

Key concepts

Convergence and Divergence

In mathematics, understanding whether a series converges or diverges is fundamental. When we talk about the convergence of a series, it means that the series approaches a specific value as more terms are added. Conversely, if a series diverges, it means that the series does not settle towards any particular value. In the context of the telescoping series we are examining, the behavior of the partial sums as the number of terms goes to infinity is crucial. For instance, if the limit of the sequence of partial sums exists and is finite, the series converges. On the other hand, if the limit is infinite or does not exist, it implies divergence. For our specific problem, as demonstrated, the limit of the nth partial sum, \(S_n\), as \(n\) approaches infinity, is \(\infty\) because the expression \(\sqrt{n+1} - \sqrt{1}\) grows without bound. Thus, this confirms that the series diverges.

Partial Sums

The concept of partial sums is essential when dealing with infinite series. A partial sum is the sum of the first n terms of a sequence. You can think of it as a snapshot of the progression of a series. In our problem, we have the sequence \(\sqrt{k+1} - \sqrt{k}\), which, when summed from 1 through \(n\), forms a telescoping series. Telescoping series are special because successive terms cancel out, making things much simpler. The partial sum \(S_n\) is simplified dramatically to just \(\sqrt{n+1} - \sqrt{1}\). This simplification happens because most of the terms in the sum cancel each other out: each term subtracts \(\sqrt{k}\) and adds \(\sqrt{k+1}\), leaving only the first and last terms uncancelled.

Limits

Limits are a vital concept in calculus, allowing us to understand the behavior of functions as inputs approach certain points. They are especially crucial when dealing with the convergence or divergence of series. To determine the behavior of our series, we evaluated the limit of the partial sum \(S_n = \sqrt{n+1} - \sqrt{1}\) as \(n\) approaches infinity. Calculating this limit involves observing that as \(n\) becomes very large, \(\sqrt{n+1}\) continues to increase indefinitely. That means the limit is \(\infty\). This result is crucial because it shows that the series does not approach a specific, finite value. Thus, the series diverges according to our limit evaluation. Understanding limits therefore provides clarity on whether a series converges or diverges by showing us the ultimate fate of the sequence of partial sums.