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. \(^{n \rightarrow \infty}\) $$\sum_{k=1}^{\infty} \ln \left(\frac{k+1}{k}\right)$$

Short answer

$$\sum_{k=1}^{\infty} \ln \left(\frac{k+1}{k}\right)$$ Answer: The given telescoping series diverges.

Step-by-step solution

Consider the first few terms of the series: $$\ln \left(\frac{2}{1}\right)+\ln \left(\frac{3}{2}\right)+\ln \left(\frac{4}{3}\right)+\cdots+\ln \left(\frac{k+1}{k}\right)$$ Notice that the terms of the series can be rewritten as differences in the logarithmic function using the formula \(\ln(a)-\ln(b)=\ln \left(\frac{a}{b}\right)\): $$\ln 2-\ln 1+\ln 3-\ln 2+\ln 4-\ln 3 \cdots+\ln(k+1)-\ln{k}$$ Now, we can find the general term for the sequence of partial sums \(S_n\) as the sum of the difference of consecutive logarithms: $$S_{n} = \sum_{k=1}^{n} \left( \ln (k+1)-\ln k \right)$$ #Step 2: Evaluate the partial sum formula#

For each \(n\), we can evaluate the sum by canceling out the common terms: \begin{align*} S_{n} &= \left(\ln 2-\cancel{\ln 1}\right)+\left(\cancel{\ln 3}-\cancel{\ln 2}\right)+\left(\cancel{\ln 4}-\cancel{\ln 3}\right)+\cdots+\left(\ln (n+1)-\cancel{\ln n}\right) \\ &= \ln(n+1)-\ln1 \\ &= \ln \left(\frac{n+1}{1}\right) \end{align*} So, the general term for the sequence of partial sums \(S_n\) is: $$S_{n} = \ln \left(\frac{n+1}{1}\right)$$ #Step 3: Find the limit of \(S_n\) as n goes to infinity#

Now that we have the formula for the nth term of the sequence of partial sums, we can find the limit as \(n\) goes to infinity: $$\lim _{n\rightarrow\infty } S_n = \lim _{n\rightarrow\infty } \ln \left(\frac{n+1}{1}\right)$$ As \(n\) goes to infinity, the expression inside the logarithm also goes to infinity. Therefore, the natural logarithm function will also approach infinity: $$\lim _{n\rightarrow\infty } S_n = \infty$$ #Step 4: State the value of the series or if it diverges#

Since the limit of the sequence of partial sums \(S_n\) goes to infinity, the original series: $$\sum_{k=1}^{\infty} \ln \left(\frac{k+1}{k}\right)$$ diverges.

Key concepts

Partial Sums

When dealing with series, a crucial concept is understanding partial sums. The partial sum, denoted as \( S_n \), represents the sum of the first \( n \) terms of a series. It helps in analyzing whether the series converges to a finite value or diverges. In our specific example, the series is expressed as:\[S_{n} = \sum_{k=1}^{n} \left( \ln (k+1) - \ln k \right)\]The partial sums of this series take advantage of the properties of logarithms, where terms cancel each other out. So for each \( n \), most terms simplify, and we are left with:\[S_{n} = \ln(n+1)\]This notation simplifies the analysis, as it gracefully highlights behavior when the number of terms increases. Recognizing this pattern in partial sums is key, and students can apply it to other similar sequences.

Logarithmic Series

A logarithmic series is formed by logarithmic expressions in its terms. In this exercise, we have the logarithmic series:\[\sum_{k=1}^{\infty} \ln \left(\frac{k+1}{k}\right)\]Here, the term \( \ln \left(\frac{k+1}{k}\right) \) can be rewritten using the logarithm properties as:\[\ln(k+1) - \ln(k)\]This representation reveals the telescoping nature of the series. Each term mostly cancels with its successive term when summed in the context of partial sums. Essentially, the series was devised by sequentially taking natural logarithms of adjacent numbers, making it shrink significantly when summed piece by piece.Understanding logarithmic series and how they telescope is vital, as it demonstrates an elegant form of simplification possible in calculus.

Series Divergence

Series divergence occurs when the partial sums of the series do not settle at a finite value as \( n \) approaches infinity. For this series, the partial sum \( S_n = \ln(n+1) \) grows without bound as \( n \) increases. Thus, its limit becomes:\[\lim _{n\rightarrow\infty } S_n = \infty\]As this limit is infinite, it signals that the series diverges. Divergence indicates that the infinite addition of terms does not converge to a specific real number, rather it increases continuously. Recognizing divergence is an important skill when studying series in calculus because it tells us that the series does not have a sum. In our case, even though the added terms become increasingly small, they accumulate to an infinite sum, showcasing the significance of analyzing series convergence or divergence correctly.