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Magazine Chapter 8: Problem 39
a. Show that \(\sum_{n=1}^{\infty} \frac{(-1)^{n}(2 n+1)}{n(n+1)}\) converges. b. Find the sum of the series of part (a).
Short Answer
Expert verified
The sum of the series \(\sum_{n=1}^{\infty} \frac{(-1)^{n}(2n+1)}{n(n+1)}\) is \(-\frac{3}{2}\).
Step by step solution
01
Write down the general term of the series
The series can be written as \(\sum_{n=1}^{\infty} \frac{(-1)^{n}(2n+1)}{n(n+1)}\). Let's denote the term by \(a_n = \frac{(-1)^{n}(2n+1)}{n(n+1)}\).
02
Apply the Alternating Series Test
To prove convergence, we will use the Alternating Series Test, which states that a series converges if:
1. \(|a_{n+1}| \le |a_n|\), and
2. \(\lim_{n\to\infty} a_n = 0\).
We notice that our series is an alternating series as each term has a factor of \((-1)^n\). Now let's apply the test to the series.
03
Check if \(|a_{n+1}| \le |a_n|\)
From the series, we can isolate the non-alternating part.
\[|a_n| = \frac{(2n+1)}{n(n+1)}\]
\[|a_{n+1}| = \frac{(2(n+1) + 1)}{(n+1)(n+2)}\]
To check if \(|a_{n+1}| \le |a_n|\), we can look at the ratio:
\[\frac{|a_{n+1}|}{|a_n|} = \frac{2n+3}{n+2} \times \frac{n(n+1)}{2n+1}\]
Since the function is decreasing, we see that the condition \(|a_{n+1}| \le |a_n|\) is met.
04
Check if \(\lim_{n\to\infty} a_n = 0\)
Now, let's check if the limit of \(a_n\) tends to 0 as n approaches infinity.
\[\lim_{n\to\infty} \frac{(-1)^{n}(2n+1)}{n(n+1)} = 0\]
Both criteria of the Alternating Series Test are satisfied, so we can conclude that the given series converges in part (a).
For part (b) 鈥� Finding the sum of the convergent series:
05
Simplify the terms in the series
Recall that the series can be written as \(\sum_{n=1}^{\infty} \frac{(-1)^{n}(2n+1)}{n(n+1)}\).
We can break the expression into two smaller fractions:
\[\frac{(2n+1)}{n(n+1)} = \frac{2n}{n(n+1)} + \frac{1}{n(n+1)} = \frac{2}{n} - \frac{1}{n+1}\]
06
Observe the telescoping property of the series
The series now becomes:
\[\sum_{n=1}^{\infty} (-1)^n \left(\frac{2}{n} - \frac{1}{n+1}\right)\]
Writing the first few terms explicitly:
\[\left(-2 +\frac{1}{2}\right) + \left( \frac{2}{2} - \frac{1}{3}\right) + \left(- \frac{2}{3} + \frac{1}{4}\right) + \left(\frac{2}{4} - \frac{1}{5}\right) + \dots\]
This is a telescoping series, meaning that some terms get canceled out in each step. In our case, the series simplifies to:
\[-2 + \frac{1}{2} + \right(\frac{2}{2} - \frac{1}{2} - \frac{1}{3}\right) + (- \frac{2}{3} + \frac{1}{3}) + \dots\]
07
Find the sum of the series
Since the series converges, we can sum all the terms that remain:
\[-2 + \frac{1}{2} = -\frac{3}{2}\]
So, the sum of the series in part (a) is \(-\frac{3}{2}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Convergent Series
Understanding a convergent series is crucial in calculus. A series is a sum of terms in a sequence. If adding an infinite number of terms results in a finite number, the series converges. In simpler terms, if you keep adding terms and the sum doesn't go to infinity, you've got a convergent series on your hands.
A convergent series means that as you add more and more terms, the total sum gets closer and closer to a specific number. This rule helps us identify convergence: a necessary condition is that the terms of the series get smaller and smaller, approaching zero.
A convergent series means that as you add more and more terms, the total sum gets closer and closer to a specific number. This rule helps us identify convergence: a necessary condition is that the terms of the series get smaller and smaller, approaching zero.
- The Alternating Series Test is a common test to determine convergence. It requires the series to have terms that alternate in sign.
- If the absolute value of the terms decreases steadily to zero, then the series converges by the test.
Telescoping Series
A telescoping series is a special type of series where consecutive terms cancel each other out. This make summing them straightforward. Telescoping series often simplify to a much easier sum, making calculations less daunting.
Picture an old-style pocket telescope that collapses in on itself. Similarly, in a telescoping series, intermediate terms vanish after canceling out their partners. What鈥檚 left over gives the series its sum.
Picture an old-style pocket telescope that collapses in on itself. Similarly, in a telescoping series, intermediate terms vanish after canceling out their partners. What鈥檚 left over gives the series its sum.
- You'll write the series in a form where most terms cancel out: usually as a difference between consecutive terms.
- This results in only a few terms that don't get canceled, significantly simplifying the calculation.
Limit of Sequence
In calculus, particularly when discussing series, the limit of a sequence is a fundamental concept. It helps in understanding if a series converges. Essentially, the limit of a sequence describes the value that the terms get closer to as the sequence progresses towards infinity.
Imagine the base of a series as a sequence of numbers. Determining the limit of this sequence gives us crucial information:
Imagine the base of a series as a sequence of numbers. Determining the limit of this sequence gives us crucial information:
- If the terms of the sequence approach zero as the sequence progresses, you鈥檙e often (with some additional criteria) dealing with a converging series.
- The limit tells us both about the behavior of the sequence over time and helps decide on the sum's convergence in a series.
