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It is useful to write series both in the form \(a_{1}+a_{2}+a_{3}+\cdots\) and in the form \(\sum_{n=1}^{\infty} a_{n} .\) Write out several terms of the following series (that is, write them in the first form). $$ \sum_{n=1}^{\infty} \frac{n}{2^{n}} $$

Short Answer

Expert verified
\(\frac{1}{2} + \frac{2}{4} + \frac{3}{8} + \frac{4}{16} + \cdots\)

Step by step solution

01

Understand the Series

This series is given in the summation notation: \ \ \ \ \ \ \ \ \ \ \(\sum_{n=1}^{\infty} \frac{n}{2^{n}}\). We need to write it in the expanded form, showing the first few terms explicitly.
02

Identify the General Term

The general term of the series is \ \ \ \(\frac{n}{2^{n}}\). For different values of \(n\), this term will change.
03

Write Several Terms

Now, substitute the first few values of \(n\) into the general term to get the series in expanded form: \ \ \ \ \[ \frac{1}{2^{1}} + \frac{2}{2^{2}} + \frac{3}{2^{3}} + \frac{4}{2^{4}} + \cdots \] \ \

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Summation Notation
Summation notation is a compact way to represent a sum of terms. Instead of writing out each term individually, summation notation groups them into a single expression. The series \(\sum_{n=1}^{\infty} \frac{n}{2^n}\) can be interpreted as: 'sum all terms where the general term is \(\frac{n}{2^n}\), starting at \(n = 1\) and continuing to infinity.' This makes it easier to write and work with large series. Understanding summation notation is crucial for dealing with mathematical series. It simplifies complex sums, making it easier to identify patterns and compute results.
General Term Identification
The general term of a series helps to find individual terms without listing them all. In the series given by \(\sum_{n=1}^{\infty} \frac{n}{2^n}\) , the general term is \( \frac{n}{2^n}\). This formula captures the nth term in the sequence. By plugging in different values for \(n\), we can find specific terms. For example, if \(n = 1\), we calculate \(\frac{1}{2^1} = \frac{1}{2}\). If \(n = 2\), then it is \(\frac{2}{2^2} = \frac{2}{4} = \frac{1}{2}\). Understanding the general term formula is critical because it allows us to see the pattern and behavior of the series as it progresses.
Series Expansion Steps
Expanding a series means writing out its terms explicitly. Following the series \( \sum_{n=1}^{\infty} \frac{n}{2^n}\), let's break it down step-by-step:
  • Step 1: Recognize the general term \(\frac{n}{2^n}\).
  • Step 2: Substitute the first few values of \(n\) into the general term:
When \(n = 1\), we get \(\frac{1}{2}\).

When \(n = 2\), we get \(\frac{2}{4} = \frac{1}{2}\).

When \(n = 3\), we get \(\frac{3}{8}\).

When \(n = 4\), we get \(\frac{4}{16} = \frac{1}{4}\).



So, the expanded form of the series is:

\( \frac{1}{2} + \frac{1}{2} + \frac{3}{8} + \frac{1}{4} + \cdots\). Breaking down the series helps in better understanding how each term is formed and how the series progresses.

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Most popular questions from this chapter

Show that it is possible to stack a pile of identical books so that the top book is as far as you tike to the right of the bottom book. You may use as many books as you need, but for any desired distance it will be some finite number. Hint: Start at the top of the pile to see how to stack them. Each time place the pile already completed on top of another book so that the pile is just at the point of tipping. (In practice, of course, you can't let them overhang quite this much without having the stack topple.) Find the distance from the right-hand end of each book to the right-hand end of the one beneath it. (To find a general formula for this distance, consider the three forces acting on book \(n\), and write the equation for the torque about its right-hand end.) Show that the sum of these distances is a divergent series. (It is said that some students studying in the library tried this experiment one evening and lefe the results, to the consternation of the librarian the next morning. It is suggested that you try it instead with one or more decks of cards. Also see "Leaning Tower of The Physical Reciens," American Journal of Physics, vol. 27, no. 2, p. 12 t.)

\(\sum_{n=1}^{\infty} \frac{x^{2 n}}{2^{n} n^{2}}\)

Prove the ratio test. Hint : If \(\left|a_{n+1} / a_{n}\right| \rightarrow \rho<1\), take \(\sigma\) so that \(\rho<\sigma<1\). Then \(\left|a_{n+1} / a_{n}\right|<\sigma\) if \(n\) is largc, say \(n \geq N\). This means that we have \(\left|a_{y-1}\right|<\sigma\left|a_{N}\right|\), \(\left|a_{N+2}\right|1\) diverges. Hint: Take \(\rho>\sigma>1\), and use the prcliminary test.

\(\sum_{n=1}^{\infty} \frac{n !}{(n+1) !}\)

It is useful to write series both in the form \(a_{1}+a_{2}+a_{3}+\cdots\) and in the form \(\sum_{n=1}^{\infty} a_{n} .\) Write out several terms of the following series (that is, write them in the first form). $$ \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n} $$

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