# A limit of a sequence \ series

• Dec 4th 2009, 02:07 AM
A limit of a sequence \ series
Prove that the sequence (why do they call it a sequence?):
$\sum_{k=n}^{3n}\frac{1}{k}$ has a limit 'a', which is in the range [ $\frac{2}{3},\frac{3}{2}$]

I don't even know how to look at it : as a series, or as a sequence. I know it should be really easy, and should use some plain lemmas or convergence tests, I just don't know which one.

Thanks!
• Dec 4th 2009, 02:37 AM
chisigma
From the well known relation...

$\lim_{n \rightarrow \infty} \sum_{k=1}^{n}\frac{1}{k} - \ln n = \gamma$ (1)

... where $\gamma$ is the so called 'Euler's constant' we derive...

$\lim_{n \rightarrow \infty} \sum_{k=1}^{3n}\frac{1}{k} - \ln 3n
=\lim_{n \rightarrow \infty} \sum_{k=1}^{n-1}\frac{1}{k} - \ln (n-1) = \gamma$
(2)

... and because is...

$\sum_{k=n}^{3n}\frac{1}{k} = \sum_{k=1}^{3n}\frac{1}{k} - \sum_{k=1}^{n-1}\frac{1}{k}$ (3)

... we have...

$\lim_{n \rightarrow \infty} \sum_{k=n}^{3n}\frac{1}{k} = \lim_{n \rightarrow \infty} \{\sum_{k=1}^{3n}\frac{1}{k} - \sum_{k=1}^{n-1}\frac{1}{k} \}=$

$\lim_{n \rightarrow \infty} \{ \ln 3n - \ln (n-1)\} = \lim_{n \rightarrow \infty} \ln \frac{3n}{n-1} = \ln 3 = 1.09861228866 ...$ (4)

Kind regards

$\chi$ $\sigma$
• Dec 4th 2009, 02:44 AM
Wow, that's amazing! We haven't learned that yet, but it looks very interesting!

But unfortunately I can't use this, I need to use more plain techniques for proving this...

Thank you!
• Dec 4th 2009, 02:52 AM
CaptainBlack
Quote:

Prove that the sequence (why do they call it a sequence?):
$\sum_{k=n}^{3n}\frac{1}{k}$ has a limit 'a', which is in the range [ $\frac{2}{3},\frac{3}{2}$]

I don't even know how to look at it : as a series, or as a sequence. I know it should be really easy, and should use some plain lemmas or convergence tests, I just don't know which one.

Thanks!

It's a sequence:

$S_n=\sum_{k=n}^{3n}\frac{1}{k}$

How you proceed depends on what you know and/or what you were doing in class at the time this was set. I would trap $S_n$ between a pair of integrals and using the squeeze theorem show that:

$\lim_{n \to \infty}S_n=\ln(3)$

CB
• Dec 4th 2009, 03:00 AM
Why is it a sequence? It looks as if $a_k=\frac{1}{k}$ is the sequence, and the sum of it is the series.
• Dec 4th 2009, 03:46 AM
CaptainBlack
Quote:

Why is it a sequence? It looks as if $a_k=\frac{1}{k}$ is the sequence, and the sum of it is the series.

The sums form a sequence.

For an infinite series the sum if it exists is the limit of the sequence of partial sums.

CB
• Dec 4th 2009, 04:02 AM
simplependulum
the lower bound is obvious :

$S = \frac{1}{n} + \frac{1}{n+1} + \frac{1}{n+2} + ... + \frac{1}{3n-1} + \frac{1}{3n}$

since

$\frac{1}{n} > \frac{1}{n+1} > ... > \frac{1}{3n}$

the series is greater than

$(2n+1) \frac{1}{3n}$

if n tends to infinity ,

$S > \frac{2}{3}$

For the upper bound ,

$S = \frac{1}{n} + \frac{1}{n+1} + \frac{1}{n+2} + ... + \frac{1}{3n-1} + \frac{1}{3n}$

$= \left ( \frac{1}{n} + \frac{1}{3n} \right ) + \left ( \frac{1}{n+1} + \frac{1}{3n-1} \right ) + ... + \left( \frac{1}{2n-1} + \frac{1}{2n+1} \right ) + \frac{1}{2n}$

$= \frac{4n}{n(3n)} + \frac{4n}{(n+1)(3n-1)} + ... + \frac{4n}{(2n-1)(2n+1)} + \frac{1}{2n}$

Consider

$(n+x)(3n-x) = -x^2 + 2nx + 3n^2$

$= -(x-n)^2 + 4n^2$

we can see that for $0 \leq x \leq n-1$

$(n+x)(3n-x)$ is min. when $x=0$

$\frac{1}{n(3n)}$ is max.

therefore , $S < (4n) \frac{n}{n(3n) } + \frac{1}{2n} = \frac{4}{3} + \frac{1}{2n}$

$\frac{4}{3}$ could be the upper bound which is better than 1.5
• Dec 4th 2009, 04:20 AM
simplependulum
Quote:

Originally Posted by chisigma
From the well known relation...

$\lim_{n \rightarrow \infty} \sum_{k=1}^{n}\frac{1}{k} - \ln n = \gamma$ (1)

... where $\gamma$ is the so called 'Euler's constant' we derive...

$\lim_{n \rightarrow \infty} \sum_{k=1}^{3n}\frac{1}{k} - \ln 3n
=\lim_{n \rightarrow \infty} \sum_{k=1}^{n-1}\frac{1}{k} - \ln (n-1) = \gamma$
(2)

... and because is...

$\sum_{k=n}^{3n}\frac{1}{k} = \sum_{k=1}^{3n}\frac{1}{k} - \sum_{k=1}^{n-1}\frac{1}{k}$ (3)

... we have...

$\lim_{n \rightarrow \infty} \sum_{k=n}^{3n}\frac{1}{k} = \lim_{n \rightarrow \infty} \{\sum_{k=1}^{3n}\frac{1}{k} - \sum_{k=1}^{n-1}\frac{1}{k} \}=$

$\lim_{n \rightarrow \infty} \{ \ln 3n - \ln (n-1)\} = \lim_{n \rightarrow \infty} \ln \frac{3n}{n-1} = \ln 3 = 1.09861228866 ...$ (4)

Kind regards

$\chi$ $\sigma$

hi

i guess you love Euler's constant very much because i can always see how you make use of it (Clapping)

$
\lim_{n \rightarrow \infty} \sum_{k=1}^{n}\frac{1}{k} - \ln n = \gamma
$

It is another special constant besides $\pi$ and $e$ . Nice indeed !
• Dec 4th 2009, 05:21 AM
Quote:

Originally Posted by simplependulum

$\frac{1}{n} > \frac{1}{n+1} > ... > \frac{1}{3n}$

the series is greater than

$(2n+1) \frac{1}{3n}$

Why is that?

And if a series is bounded, then it's still not enough for saying that it has a limit.
• Dec 4th 2009, 05:52 PM
simplependulum
Quote:

Why is that?

And if a series is bounded, then it's still not enough for saying that it has a limit.

I could just explain your first question , other is left to some experts

$S = \frac{1}{n} + \frac{1}{n+1} + \frac{1}{n+2} + ... + \frac{1}{3n-1} + \frac{1}{3n}$

we know that there is $3n - (n) + 1 = 2n+1$ terms here .

and since $\frac{1}{n} > \frac{1}{n+1} > \frac{1}{n+2} >... > \frac{1}{3n}$

therefore

$S > \frac{1}{3n} + \frac{1}{3n} + ... + \frac{1}{3n}$

There are totally $2n+1$ terms so

$S > (2n+1)\frac{1}{3n}$
• Dec 4th 2009, 06:09 PM
Krizalid
for each $n\le k\le3n$ we have $\frac{1}{3n}\le \frac{1}{k}\le \frac{1}{n},$ hence $\sum\limits_{k=n}^{3n}{\frac{1}{3n}}\le \sum\limits_{k=n}^{3n}{\frac{1}{k}}\le \sum\limits_{k=n}^{3n}{\frac{1}{n}},$ and $\frac{2n+1}{3n}\le \sum\limits_{k=n}^{3n}{\frac{1}{k}}\le \frac{2n+1}{n},$ finally as $n\longrightarrow\infty$ you'll get the lower bound $\frac23$ and upper bound $2,$ which is a better result that you asked!
• Dec 4th 2009, 06:23 PM
simplependulum
Quote:

Originally Posted by Krizalid
for each $n\le k\le3n$ we have $\frac{1}{3n}\le \frac{1}{k}\le \frac{1}{n},$ hence $\sum\limits_{k=n}^{3n}{\frac{1}{3n}}\le \sum\limits_{k=n}^{3n}{\frac{1}{k}}\le \sum\limits_{k=n}^{3n}{\frac{1}{n}},$ and $\frac{2n+1}{3n}\le \sum\limits_{k=n}^{3n}{\frac{1}{k}}\le \frac{2n+1}{n},$ finally as $n\longrightarrow\infty$ you'll get the lower bound $\frac23$ and upper bound $2,$ which is a better result that you asked!

Hi ,

adam63 asked how to prove there exists a limit which is bounded . I think it is obvious but i do not know how to complete this proof ...
• Dec 5th 2009, 01:31 AM
CaptainBlack
Quote:

Originally Posted by Krizalid
for each $n\le k\le3n$ we have $\frac{1}{3n}\le \frac{1}{k}\le \frac{1}{n},$ hence $\sum\limits_{k=n}^{3n}{\frac{1}{3n}}\le \sum\limits_{k=n}^{3n}{\frac{1}{k}}\le \sum\limits_{k=n}^{3n}{\frac{1}{n}},$ and $\frac{2n+1}{3n}\le \sum\limits_{k=n}^{3n}{\frac{1}{k}}\le \frac{2n+1}{n},$ finally as $n\longrightarrow\infty$ you'll get the lower bound $\frac23$ and upper bound $2,$ which is a better result that you asked!

The upper bound asker for is 3/2 which is tighter than 2.

CB
• Dec 5th 2009, 03:30 AM
Krizalid
ah yes, i got my mind in clouds. :D