Question

Find the sum of the series $$ \frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\frac{1}{4 \cdot 5}+\cdots $$ Hint : \(\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}\) Show that the remainder after \(n\) terms is \(1 /(n+1)\). Hence show that about 200 terms are needed for two decimal place accuracy. Compare the remainder with the value of the 200th term and so show that in computation using series of positive terms the value of the first omitted term may be a completely unreliable estimate of the error.

Short answer

200 terms are needed for two decimal place accuracy. Remainder is 0.00498. Using first omitted term as error can be unreliable.

Step-by-step solution

Substitute the Hint

Using the hint provided \[\begin{equation}\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}\end{equation}\] We can rewrite the given series as \[\begin{equation}\sum_{n=1}^\infty \frac{1}{n(n+1)} = \sum_{n=1}^\infty \left( \frac{1}{n} - \frac{1}{n+1} \right)\end{equation}\]

Observe the Telescoping Series

Notice that this is a telescoping series where most of the terms cancel out. For example, the first few terms are:\[\begin{equation}(\frac{1}{1} - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + \ldots\end{equation}\] Many intermediate terms cancel out.

Sum of the First n Terms

By evaluating the first n terms of the telescoping series, we see the cancellation effect:\[\begin{equation}S_n = \left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \cdots + \left( \frac{1}{n} - \frac{1}{n+1} \right) = 1 - \frac{1}{n+1}\end{equation}\]

Remainder After n Terms

From Step 3, the sum of the first n terms is \[\begin{equation}S_n = 1 - \frac{1}{n+1}\end{equation}\]. Hence, the remainder after n terms is \[\begin{equation}R_n = \frac{1}{n+1}\end{equation}\]. To verify this for two decimal place accuracy, set \[\begin{equation}R_n \leq 0.005\end{equation}\]. Solving \[\begin{equation}\frac{1}{n+1} \leq 0.005, \text{ we find that } n \geq 199.\end{equation}\]

Comparing Remainder with 200th Term

The remainder after 200 terms is: \[\begin{equation}R_{200} = \frac{1}{201} \approx 0.00498\end{equation}\]. The 200th term, given by \[\begin{equation}T_{200} = \frac{1}{200 \cdot 201}\end{equation}\], is significantly smaller than the remainder. Therefore, it shows that in a telescoping series, using the first omitted term as an error might lead to inaccurate estimates.

Key concepts

telescoping series

In mathematics, a telescoping series is a series whose partial sums eventually only include a fixed number of terms after cancellation. This means that for a given series, many terms add to zero and thus cancel each other out. This property makes telescoping series quite useful in simplifying complex series into simpler expressions.

series summation

Series summation is the process of adding the terms of a series to find their total sum. In the case of a telescoping series, this process is simplified because of the cancellation effect between terms. For the given series \(\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}\), we observe this effect clearly:

• The first term is \(1 - \frac{1}{2}\)
• The second term is \( \frac{1}{2} - \frac{1}{3}\)
• Continuing this pattern, intermediate terms cancel out
• The remaining expression is \(1 - \frac{1}{n+1}\)

This reduces the problem to a manageable form which makes it easier to find the sum.

error estimation

Error estimation determines how close the sum of the partial series is to the actual sum of the infinite series. For the given series, the remainder after 'n' terms is given by \(\frac{1}{n+1}\).To find the number of terms needed for accuracy, solve \( \frac{1}{n+1} \leq 0.005 \).

• After solving, we find \( n \geq 199 \)
• This means 200 terms are needed for the sum to be accurate to two decimal places

It is much more efficient and tells how many terms we need to achieve a specific accuracy.

mathematical convergence

Mathematical convergence refers to the convergence of a series as the number of terms approaches infinity. For a series to converge, the sequence of partial sums must approach a finite limit. In the provided exercise, the series converges because as \( n \to \infty\), the remainder term \( \frac{1}{n+1} \) approaches zero.

• Thus, \(\frac{1}{n+1}\) makes the series sum converge to a finite value
• It also highlights the concept that the first omitted term might not be a reliable error estimate

Understanding convergence helps predict the behavior of series as terms increase, offering insights into whether a series will sum to a finite number.