sequences/series - just a silly question :)

[DivGradCurl]

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.
\lim _{n \to \infty} \frac{1}{n} = 0

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

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


2.
\lim _{n \to \infty} \frac{n}{n+1} = \lim _{n \to \infty} \frac{1}{1+\frac{1}{n}}=1

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

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

[HallsofIvy]

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.

[DivGradCurl]

I get your point. So, I've just got partial sums to do the job... right?

[Zurtex]

Never done sums as nasty as that yet but the second one is really is easy to approach. Remember that:

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

One you get that simply work out \lim_{r \rightarrow \infty} and you'll see whether or not it converges and if so what value you get (done the workings it is quite easy).

[NateTG]

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.
\lim _{n \to \infty} \frac{1}{n} = 0

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

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


2.
\lim _{n \to \infty} \frac{n}{n+1} = \lim _{n \to \infty} \frac{1}{1+\frac{1}{n}}=1

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

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

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.

[DivGradCurl]

\textrm{Guys, this is where I could get to:}

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.}

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

\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.}

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.}

\textrm{Thanks}

[NateTG]

\textrm{Guys, this is where I could get to:}

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.}


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:
\sum_{n=1}^{\infty} (n - n) = \sum_{n=1}^{\infty} 0 = 0
but according to your logic
\sum_{n=1}^{\infty} (n - n) \rightarrow \sum_{n=1}^{\infty} n - \sum_{n=1}^{\infty}n = \infty-\infty

Perhaps you could write out the first two or three terms of the series, and look for a pattern?

[DivGradCurl]

\textrm{How about the following?}

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

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)

\textrm{gives the partial sums}

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

\textrm{Therefore,}

\sum _{n = 1} ^{\infty} \ln \left( n \right) - \ln \left( n +1 \right) = -\infty \Longrightarrow \textrm{The series diverges.}

[NateTG]

Those answers are probably acceptable, but you might want to clean up your nontation a bit:

I would have said that:
\frac{1}{2n} < sin \frac{1}{n}
and
\sum_n^\infty \frac{1}{2n}
is divergent. So
\sum_n^\infty \frac{1}{2n}
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
s_n=-ln(n+1), 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:
\sum_{n=i}^m (f(n+1)-f(n))
clearly has the formula
\sum_{n=i}^m (f(n+1)-f(n))=f(m+1)-f(i)
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.

[DivGradCurl]

Thanks a lot.