Sequences/series - just a silly question

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In summary, the conversation discusses finding the values of series involving sine, cosine, or ln functions, and the general approach to solving them. The first series diverges and the second series also diverges due to it being a telescoping sum. The conversation also includes tips on notation and writing out partial sums.
  • #1
DivGradCurl
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sequences/series - just a silly question :)

I think the limits I have found for the sequences below are right, but I'm not sure on how to approach the series. Please, help me understand what I should do to find the series when a sine, cosine, or a ln appears in the formula. Is there a general procedure?

Thank you very much.
:smile:

1.
[tex] \lim _{n \to \infty} \frac{1}{n} = 0 [/tex]

[tex] \lim _{n \to \infty} \sin \left( \frac{1}{n} \right) = \sin 0 = 0 [/tex]

[tex] \sum _{n = 1} ^{\infty} \sin \left( \frac{1}{n} \right) = ? [/tex]


2.
[tex] \lim _{n \to \infty} \frac{n}{n+1} = \lim _{n \to \infty} \frac{1}{1+\frac{1}{n}}=1 [/tex]

[tex] \lim _{n \to \infty} \ln \left( \frac{n}{n+1} \right) = \ln 1 = 0 [/tex]

[tex] \sum _{n = 1} ^{\infty} \ln \left( \frac{n}{n+1} \right) = ? [/tex]
 
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  • #2
Where did you get these problems? If you had got a limit other than 0 for the sequence then you would know that the series does not converge. However, the fact that the limit is 0 does not tell you that the series does converge, much less what the limit might be if it does.
 
  • #3
I get your point. So, I've just got partial sums to do the job... right?
 
  • #4
Never done sums as nasty as that yet but the second one is really is easy to approach. Remember that:

[tex]\sum_{n=1}^{r} \left( \ln \frac{n}{n+1} \right) = \sum_{n=1}^{r} \left( \ln (n) - \ln (n+1)\right)[/tex]

One you get that simply work out [itex]\lim_{r \rightarrow \infty}[/itex] and you'll see whether or not it converges and if so what value you get (done the workings it is quite easy).
 
  • #5
thiago_j said:
I think the limits I have found for the sequences below are right, but I'm not sure on how to approach the series. Please, help me understand what I should do to find the series when a sine, cosine, or a ln appears in the formula. Is there a general procedure?

Thank you very much.
:smile:

1.
[tex] \lim _{n \to \infty} \frac{1}{n} = 0 [/tex]

[tex] \lim _{n \to \infty} \sin \left( \frac{1}{n} \right) = \sin 0 = 0 [/tex]

[tex] \sum _{n = 1} ^{\infty} \sin \left( \frac{1}{n} \right) = ? [/tex]


2.
[tex] \lim _{n \to \infty} \frac{n}{n+1} = \lim _{n \to \infty} \frac{1}{1+\frac{1}{n}}=1 [/tex]

[tex] \lim _{n \to \infty} \ln \left( \frac{n}{n+1} \right) = \ln 1 = 0 [/tex]

[tex] \sum _{n = 1} ^{\infty} \ln \left( \frac{n}{n+1} \right) = ? [/tex]

Aside from numerical methods, there is no general approach to finding the values of series. Zurtex has essentially given away the second one already.

To get an answer to the first one, you might consider comparing it to a different series that you should already be familiar with.
 
  • #6
[tex] \textrm{Guys, this is where I could get to:} [/tex]

[tex]1.~ \sum _{n = 1} ^{\infty} \sin \left( \frac{1}{n} \right) < \sum _{n = 1} ^{\infty} \frac{1}{n} \Longrightarrow \textrm{The comparison to the harmonic series is not conclusive.} [/tex]

[tex] \textrm{(I don't know what series could be useful at this time. Any tips?)}[/tex]

[tex] \sum _{n = 1} ^{\infty} \sin \left( \frac{1}{n} \right) \Longrightarrow \int _1 ^{\infty} \sin \left( \frac{1}{x} \right) \textrm{d}x \approx 33 \quad \textrm{(converges)} \Longrightarrow \textrm{The series converges by \textit{the integral test}---partial sums give a finite answer.} [/tex]

[tex]2.~ \sum _{n = 1} ^{\infty} \ln \left( \frac{n}{n+1} \right) = \sum _{n = 1} ^{\infty} \ln \left( n \right) - \sum _{n = 1} ^{\infty} \ln \left( n +1 \right) = \infty - \infty \quad \textrm{(undefined)} \Longrightarrow \textrm{The series diverges.} [/tex]

[tex] \textrm{Thanks} [/tex]
 
  • #7
thiago_j said:
[tex] \textrm{Guys, this is where I could get to:} [/tex]

[tex]1.~ \sum _{n = 1} ^{\infty} \sin \left( \frac{1}{n} \right) < \sum _{n = 1} ^{\infty} \frac{1}{n} \Longrightarrow \textrm{The comparison to the harmonic series is not conclusive.} [/tex]

Regarding part 1:
You're really close. Hint in white try 1/(2n) instead

Regarding part 2:
It's not safe to rearange terms in conditionally convergent series, so you can't do that.
For example:
[tex]\sum_{n=1}^{\infty} (n - n) = \sum_{n=1}^{\infty} 0 = 0[/tex]
but according to your logic
[tex]\sum_{n=1}^{\infty} (n - n) \rightarrow \sum_{n=1}^{\infty} n - \sum_{n=1}^{\infty}n = \infty-\infty[/tex]

Perhaps you could write out the first two or three terms of the series, and look for a pattern?
 
  • #8
[tex] \textrm{How about the following?} [/tex]

[tex]1.~ \sum _{n = 1} ^{\infty} \sin \left( \frac{1}{n} \right) > \sum _{n = 1} ^{\infty} \frac{1}{2n} \Longrightarrow \textrm{The series diverges.} [/tex]

[tex]2.~ \sum _{n = 1} ^{\infty} \ln \left( \frac{n}{n+1} \right) = \sum _{n = 1} ^{\infty} \ln \left( n \right) - \ln \left( n +1 \right) [/tex]

[tex] \textrm{gives the partial sums} [/tex]

[tex] s_1 = -\ln (2) [/tex]
[tex] s_2 = -\ln (3) [/tex]
[tex] s_3 = -2\ln (2) [/tex]
[tex] s_4 = -\ln (5) [/tex]
[tex] s_5 = -\ln (6) [/tex]
[tex] s_6 = -\ln (7) [/tex]
[tex] s_7 = -3\ln (2) [/tex]
[tex] \vdots [/tex]
[tex] s = \lim _{n \to \infty} s_n = -\infty. [/tex]

[tex] \textrm{Therefore,} [/tex]

[tex] \sum _{n = 1} ^{\infty} \ln \left( n \right) - \ln \left( n +1 \right) = -\infty \Longrightarrow \textrm{The series diverges.} [/tex]
 
  • #9
Those answers are probably acceptable, but you might want to clean up your nontation a bit:

I would have said that:
[tex]\frac{1}{2n} < sin \frac{1}{n}[/tex]
and
[tex]\sum_n^\infty \frac{1}{2n}[/tex]
is divergent. So
[tex]\sum_n^\infty \frac{1}{2n}[/tex]
is divergent by the comparison test.
Rather than writing > with the sums, because, since the sums are divergent, it's not at all clear that > is meaningful.

For the second one, it would have been a good idea to write out that
[tex]s_n=-ln(n+1)[/tex], and it would probably also be clearer to write the numbers out rather than pulling exponents the way you did.

The second sum is an example of what is called a telescoping sum (since it collapses like a telescope).
Any sum of the form:
[tex]\sum_{n=i}^m (f(n+1)-f(n))[/tex]
clearly has the formula
[tex]\sum_{n=i}^m (f(n+1)-f(n))=f(m+1)-f(i)[/tex]
since, if it is written out, all of the terms except for the first and last will cancel pairwise.

BTW: It's usuall a good idea to put parentheses around the expression of a series if it contains + or -, it makes for much easier reading, and may prevent errors.
 
  • #10
Thanks a lot.
 

1. What is a sequence or series?

A sequence is a list of numbers or terms that follow a specific pattern or rule. A series is the sum of terms in a sequence.

2. What is the difference between an arithmetic and geometric sequence/series?

In an arithmetic sequence, each term is found by adding a constant number, called the common difference, to the previous term. In a geometric sequence, each term is found by multiplying the previous term by a constant number, called the common ratio. The difference between an arithmetic and geometric series is that in an arithmetic series, the terms are added together, while in a geometric series, the terms are multiplied together.

3. How do I find the sum of a finite arithmetic or geometric series?

The formula for finding the sum of a finite arithmetic series is Sn = (n/2)(a + L), where n is the number of terms, a is the first term, and L is the last term. The formula for finding the sum of a finite geometric series is Sn = a(1-r^n)/(1-r), where a is the first term, r is the common ratio, and n is the number of terms.

4. What is the difference between a convergent and divergent series?

A series is convergent if the sequence of partial sums approaches a finite number as the number of terms increases. A series is divergent if the sequence of partial sums does not approach a finite number and instead goes to infinity.

5. What are some real-world applications of sequences and series?

Sequences and series have many real-world applications, such as in financial investments, population growth, and radioactive decay. They are also used in calculus to approximate functions and in computer algorithms for data compression and encryption.

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