Question

Let \(S_{n}\) be the \(n^{\text {th }}\) partial sum of a series. In Exercises \(21-24,\) a convergent alternating series is given and a value of \(n .\) Compute \(S_{n}\) and \(S_{n+1}\) and use these values to find bounds on the sum of the series. $$\sum_{n=1}^{\infty} \frac{(-1)^{n}}{\ln (n+1)}, \quad n=5$$

Short answer

The sum is between -1.2841 and -0.7702.

Step-by-step solution

Identify the given series and determine terms

The given series is \(\sum_{n=1}^{\infty} \frac{(-1)^{n}}{\ln (n+1)}\). This is an alternating series. For \(n=5\), we will consider the first five terms: \(a_1 = \frac{-1}{\ln(2)}\), \(a_2 = \frac{1}{\ln(3)}\), \(a_3 = \frac{-1}{\ln(4)}\), \(a_4 = \frac{1}{\ln(5)}\), and \(a_5 = \frac{-1}{\ln(6)}\).

Compute the 5th partial sum, \(S_5\)

Calculate the partial sum, \(S_5\), which is the sum of the first 5 terms: \[S_5 = a_1 + a_2 + a_3 + a_4 + a_5 = \frac{-1}{\ln(2)} + \frac{1}{\ln(3)} + \frac{-1}{\ln(4)} + \frac{1}{\ln(5)} + \frac{-1}{\ln(6)}\].

Compute the 6th partial sum, \(S_6\)

Find \(S_6\) by adding the next term, \(a_6 = \frac{1}{\ln(7)}\), to \(S_5\): \[S_6 = S_5 + a_6 = S_5 + \frac{1}{\ln(7)}\].

Calculate specified terms and sums numerically

Evaluate each term numerically using a calculator and compute \(S_5\) and \(S_6\). \[a_1 \approx -1.4427, \quad a_2 \approx 0.9102, \quad a_3 \approx -0.7213, \quad a_4 \approx 0.6213, \quad a_5 \approx -0.6516\]Then,\[S_5 \approx -1.4427 + 0.9102 - 0.7213 + 0.6213 - 0.6516 \approx -1.2841\]For \(a_6\), \[a_6 \approx 0.5139\] \[S_6 \approx -1.2841 + 0.5139 \approx -0.7702\].

Estimate the bounds for the sum of the series

The error bound for a convergent alternating series is less than the absolute value of the first neglected term (\(|a_{6}|\)). Thus, the sum of the series, \(S\), is between \(S_5 - |a_6|\) and \(S_5 + |a_6|\) or \(S_5 < S < S_6\). Therefore, the sum \(S\) is in the range,\[-1.2841 < S < -0.7702\].

Key concepts

Partial Sums

When dealing with series, especially alternating series, understanding partial sums is crucial. A partial sum, denoted as \(S_{n}\), is the sum of the first \(n\) terms of a series. It gives us an approximation of the total sum of the entire series by adding up its initial terms.

For instance, consider the series \(\sum_{n=1}^{\infty} \frac{(-1)^{n}}{\ln(n+1)}\). To find the 5th partial sum, \(S_5\), we add together the first five terms: \[S_5 = a_1 + a_2 + a_3 + a_4 + a_5 = \frac{-1}{\ln(2)} + \frac{1}{\ln(3)} + \frac{-1}{\ln(4)} + \frac{1}{\ln(5)} + \frac{-1}{\ln(6)}\].

By calculating each term individually and summing them, we obtain the value of the 5th partial sum, which in this case is approximately \(-1.2841\). This gives an estimate of the series up to the 5th term. Adding an additional term gives us the next partial sum, \(S_6\), which provides a closer approximation to the actual value of the series. Understanding partial sums is vital as they play a key role in estimating and understanding the behavior of a series.

Convergent Series

A convergent series is a series whose partial sums tend to a finite limit as the number of terms increases indefinitely. In simpler words, as we include more and more terms in our partial sums, the total sum approaches a specific value, instead of diverging to infinity or remaining undefined.

Alternating series, such as the one given, are often convergent. This means that even though its terms are positive and negative alternatingly, the overall summation still settles on a certain value. The convergence property is very helpful for infinite series that follow a clear pattern.

For the series \(\sum_{n=1}^{\infty} \frac{(-1)^{n}}{\ln(n+1)}\), it is proved that it converges as more terms are taken into account. By calculating the partial sums \(S_5\) and \(S_6\), one can observe how the series is indeed getting closer to a specific range, ensuring it doesn't diverge into an undefined infinity. This convergence helps in making reasonable approximations for the total sum.

Error Bound

The error bound in an alternating series gives us valuable insight into how precise our approximation of the total sum is, based on the calculated partial sums. For alternating series, the error bound after a certain number of terms is less than the absolute value of the first neglected term.

In our example, after calculating the 5th partial sum \(S_5\) and the 6th term \(a_6\), the error bound can be determined by the absolute value of \(a_6\). If \(a_6 = \frac{1}{\ln(7)} \approx 0.5139\), the error bound tells us that the true sum \(S\) of the series lies between \(S_5 - 0.5139\) and \(S_5 + 0.5139\). This translates to \(-1.2841 < S < -0.7702\).

In simpler terms, the error bound provides a range in which the actual sum of the series can be found. Understanding and calculating error bounds is crucial for handling series, especially in finding how accurate our approximations are and refining our calculations if necessary.