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Limit of sum

Pranav

Well-known member
Nov 4, 2013
428
Problem:
Find the value of
$$\lim_{n\rightarrow \infty} \sum_{r=0}^n \left(\frac{1}{4r+1}-\frac{1}{4r+3}\right)$$

Attempt:
I tried writing down a few terms to see if the terms cancel but no luck there. I couldn't find any closed form for the summation. :(

Next, I thought of converting it into a definite integral. The usual approach is to consider $r/n$ as $x$ and $1/n$ as dx but I am unable to find a way to do this.

I am completely clueless now. (Doh)

Any help is appreciated. Thanks!
 
Last edited:

chisigma

Well-known member
Feb 13, 2012
1,704
Problem:
Find the value of
$$\lim_{n\rightarrow \infty} \sum_{r=0}^n \left(\frac{1}{4r+1}-\frac{1}{4r+3}\right)$$

Attempt:4\ r + 3
I tried writing down a few terms to see if the terms cancel but no luck there. I couldn't find any closed form for the summation. :(

Next, I thought of converting it into a definite integral. The usual approach is to consider $r/n$ as $x$ and $1/n$ as dx but I am unable to find a way to do this.

I am completely clueless now. (Doh)

Any help is appreciated. Thanks!
The first step is to transform Your series in...


$\displaystyle S = \sum_{r=0}^{\infty} (\frac{1}{4\ r + 1} - \frac{1}{4\ r + 3}) = \frac{1}{8}\ \sum_{r=0}^{\infty} \frac{1}{(r + \frac{1}{4})\ (r+ \frac{3}{4})}\ (1) $

... and the second step is to apply the (5.32) of...


http://mathhelpboards.com/discrete-...ation-tutorial-draft-part-i-426.html#post2494


$\displaystyle \sum_{r=1}^{\infty} \frac{1}{(r + a)\ (r + b)} = \frac{\phi(b) - \phi(a)}{b - a}\ (2)$

... where...


$\displaystyle \phi (x) = \frac{d}{dx} \ln x!\ (3)$

... so that we obtain...

$\displaystyle S = \frac{2}{3} + \frac{1}{4}\ \{\phi(\frac{3}{4}) - \phi(\frac{1}{4})\}\ (4)$

The successive step is numerical evaluation of (4) and it will be performed in next post...

Kind regards

$\chi$ $\sigma$
 

chisigma

Well-known member
Feb 13, 2012
1,704
The first step is to transform Your series in...


$\displaystyle S = \sum_{r=0}^{\infty} (\frac{1}{4\ r + 1} - \frac{1}{4\ r + 3}) = \frac{1}{8}\ \sum_{r=0}^{\infty} \frac{1}{(r + \frac{1}{4})\ (r+ \frac{3}{4})}\ (1) $

... and the second step is to apply the (5.32) of...


http://mathhelpboards.com/discrete-...ation-tutorial-draft-part-i-426.html#post2494


$\displaystyle \sum_{r=1}^{\infty} \frac{1}{(r + a)\ (r + b)} = \frac{\phi(b) - \phi(a)}{b - a}\ (2)$

... where...


$\displaystyle \phi (x) = \frac{d}{dx} \ln x!\ (3)$

... so that we obtain...

$\displaystyle S = \frac{2}{3} + \frac{1}{4}\ \{\phi(\frac{3}{4}) - \phi(\frac{1}{4})\}\ (4)$

The successive step is numerical evaluation of (4) and it will be performed in next post...
Using 'Monster Wolfram' we find the [quite surprising...] result ...

$\displaystyle \phi(\frac{3}{4}) = \frac{4}{3} - \gamma + \frac{\pi}{2} - \ln 8$

$\displaystyle \phi(\frac{1}{4}) = 4 - \gamma - \frac{\pi}{2} - \ln 8$

... so that is...

$\displaystyle S = \frac{2}{3} + \frac{1}{4}\ (\pi - \frac{8}{3}) = \frac{\pi}{4}$

... and that suggests that a more confortable way exists to arrive at the result...


Kind regards


$\chi$ $\sigma$
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Problem:
Find the value of
$$\lim_{n\rightarrow \infty} \sum_{r=0}^n \left(\frac{1}{4r+1}-\frac{1}{4r+3}\right)$$

Attempt:
I tried writing down a few terms to see if the terms cancel but no luck there. I couldn't find any closed form for the summation. :(

Next, I thought of converting it into a definite integral. The usual approach is to consider $r/n$ as $x$ and $1/n$ as dx but I am unable to find a way to do this.

I am completely clueless now. (Doh)

Any help is appreciated. Thanks!
We can use the digamma function

We can regularize a divergent infinite sum to give it a finite value

\(\displaystyle \sum_{n\geq 0}\frac{1}{n+a}=-\psi(a)\)

So

\(\displaystyle S=\frac{1}{4}\sum_{r=0}^\infty \left(\frac{1}{r+\frac{1}{4}}-\frac{1}{r+\frac{3}{4}}\right)=-\frac{1}{4} \left(\psi\left(\frac{1}{4}\right) -\psi\left(\frac{3}{4}\right)\right)\)

Now use the reflection formula

\(\displaystyle \psi(1-x)-\psi(x)=\pi\cot(\pi x)\)

so

\(\displaystyle \psi\left(\frac{1}{4}\right) -\psi \left(\frac{3}{4}\right) =-\pi\cot \left(\frac{\pi}{4} \right)=-\pi\)

Hence

\(\displaystyle S=\frac{\pi}{4}\)

- - - Updated - - -

Interestingly using the regulaized representation we can find that

\(\displaystyle \sum_{n\geq 0}\frac{1}{n+1}=\sum_{n\geq 1}\frac{1}{n}=-\psi(1)=\gamma\)

So

\(\displaystyle \lim_{n \to \infty }H_n = \gamma\)
 

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
$$ \sum_{r=0}^{\infty} \Big( \frac{1}{4r+1} - \frac{1}{4r+3} \Big) = \sum_{r=0}^{\infty} \int_{0}^{1} \Big( x^{4r} -x^{4r+2} \Big) \ dx$$

$$= \int_{0}^{1} \sum_{r=0}^{\infty} \Big(x^{4r}-x^{4r+2} \Big) = \int_{0}^{1} \frac{1-x^{2}}{1-x^{4}} \ dx = \int_{0}^{1} \frac{1}{1+x^{2}} \ dx = \frac{\pi}{4}$$
 

Pranav

Well-known member
Nov 4, 2013
428
Hi chisigma! :)

The first step is to transform Your series in...


$\displaystyle S = \sum_{r=0}^{\infty} (\frac{1}{4\ r + 1} - \frac{1}{4\ r + 3}) = \frac{1}{8}\ \sum_{r=0}^{\infty} \frac{1}{(r + \frac{1}{4})\ (r+ \frac{3}{4})}\ (1) $

... and the second step is to apply the (5.32) of...


http://mathhelpboards.com/discrete-...ation-tutorial-draft-part-i-426.html#post2494


$\displaystyle \sum_{r=1}^{\infty} \frac{1}{(r + a)\ (r + b)} = \frac{\phi(b) - \phi(a)}{b - a}\ (2)$

... where...


$\displaystyle \phi (x) = \frac{d}{dx} \ln x!\ (3)$

... so that we obtain...

$\displaystyle S = \frac{2}{3} + \frac{1}{4}\ \{\phi(\frac{3}{4}) - \phi(\frac{1}{4})\}\ (4)$

The successive step is numerical evaluation of (4) and it will be performed in next post...

Kind regards

$\chi$ $\sigma$
That looks way too complicated for me to comprehend. I have never dealt with anything like whats shown in your tutorial and it isn't even my coursework too. :confused:

Is there no other way to solve this problem? I doubt I need to go through such complicated stuff for this problem because it is an exam problem.
 

Pranav

Well-known member
Nov 4, 2013
428
We can use the digamma function

We can regularize a divergent infinite sum to give it a finite value

\(\displaystyle \sum_{n\geq 0}\frac{1}{n+a}=-\psi(a)\)

So

\(\displaystyle S=\frac{1}{4}\sum_{r=0}^\infty \left(\frac{1}{r+\frac{1}{4}}-\frac{1}{r+\frac{3}{4}}\right)=-\frac{1}{4} \left(\psi\left(\frac{1}{4}\right) -\psi\left(\frac{3}{4}\right)\right)\)

Now use the reflection formula

\(\displaystyle \psi(1-x)-\psi(x)=\pi\cot(\pi x)\)

so

\(\displaystyle \psi\left(\frac{1}{4}\right) -\psi \left(\frac{3}{4}\right) =-\pi\cot \left(\frac{\pi}{4} \right)=-\pi\)

Hence

\(\displaystyle S=\frac{\pi}{4}\)

- - - Updated - - -

Interestingly using the regulaized representation we can find that

\(\displaystyle \sum_{n\geq 0}\frac{1}{n+1}=\sum_{n\geq 1}\frac{1}{n}=-\psi(1)=\gamma\)

So

\(\displaystyle \lim_{n \to \infty }H_n = \gamma\)
Thank you for your participation ZaidAlyafey but unfortunately that's no better, it still requires a level of Math I haven't reached. :(

That's out of my coursework, I haven't yet seen the gamma function and you ask me to look for digamma function. :p

$$ \sum_{r=0}^{\infty} \Big( \frac{1}{4r+1} - \frac{1}{4r+3} \Big) = \sum_{r=0}^{\infty} \int_{0}^{1} \Big( x^{4r} -x^{4r+2} \Big) \ dx$$

$$= \int_{0}^{1} \sum_{r=0}^{\infty} \Big(x^{4r}-x^{4r+2} \Big) = \int_{0}^{1} \frac{1-x^{2}}{1-x^{4}} \ dx = \int_{0}^{1} \frac{1}{1+x^{2}} \ dx = \frac{\pi}{4}$$
Hi Random Variable!

Sorry if this is obvious but how do you convert the fractions into powers of x? :confused:

Thank you!
 

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
$$ \int_{0}^{1} x^{4r} = \frac{x^{4r+1}}{4r+1} \Big|^{1}_{0} = \frac{1}{4r+1}$$

$$ \int_{0}^{1} x^{4r+2} = \frac{x^{4r+3}}{4r+3} \Big|^{1}_{0} = \frac{1}{4r+3}$$
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Actually it might be enough to reckognize that

\(\displaystyle \sum_{n\geq 0}\frac{1}{4r+1}-\frac{1}{4r+3}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots =\sum_{n\geq 0}\frac{(-1)^n}{2n+1}=\frac{\pi}{4}\)

Where we use the series representation of \(\displaystyle \arctan(x)=\sum_{n\geq 0}\frac{(-1)^n}{2n+1}x^{2n+1}\)

by putting \(\displaystyle x=1\).
 

Pranav

Well-known member
Nov 4, 2013
428
$$ \int_{0}^{1} x^{4r} = \frac{x^{4r+1}}{4r+1} \Big|^{1}_{0} = \frac{1}{4r+1}$$

$$ \int_{0}^{1} x^{4r+2} = \frac{x^{4r+3}}{4r+3} \Big|^{1}_{0} = \frac{1}{4r+3}$$
That's very cool! :cool:

This trick looks very useful, thanks a lot Random Variable! :D

Do you know where can I find similar problems (book or a link)? I feel that the method you have shown might help me again somewhere so I want to try a few more problems, thank you! :)

Actually it might be enough to reckognize that

\(\displaystyle \sum_{n\geq 0}\frac{1}{4r+1}-\frac{1}{4r+3}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots =\sum_{n\geq 0}\frac{(-1)^n}{2n+1}=\frac{\pi}{4}\)

Where we use the series representation of \(\displaystyle \arctan(x)=\sum_{n\geq 0}\frac{(-1)^n}{2n+1}x^{2n+1}\)

by putting \(\displaystyle x=1\).
Yes, I see it. I am sure the series representation would have never have occurred to me because I am much more used to dealing with series of sines and cosines. Thank you ZaidAlyafey! :)
 

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
Unfortunately I don't have a link to similar problems.

But here's another simple example.

Somehow this didn't come up in a challenge thread about ways to evaluate the alternating harmonic series.

$$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} = \sum_{n=1}^{\infty} (-1)^{n+1} \int_{0}^{1} x^{n-1} \ dx = \int_{0}^{1} \sum_{n=1}^{\infty} (-1)^{n+1} x^{n-1} \ dx$$

$$ = \int_{0}^{1} \sum_{n=0}^{\infty} (-1)^{n+2} x^{n} = \int_{0}^{1} \sum_{n=0}^{\infty} (-1)^{n} x^{n} \ dx = \int_{0}^{1}\frac{1}{1+x} \ dx = \ln 2$$
 

Pranav

Well-known member
Nov 4, 2013
428
Unfortunately I don't have a link to similar problems.

But here's another simple example.

Somehow this didn't come up in a challenge thread about ways to evaluate the alternating harmonic series.

$$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} = \sum_{n=1}^{\infty} (-1)^{n+1} \int_{0}^{1} x^{n-1} \ dx = \int_{0}^{1} \sum_{n=1}^{\infty} (-1)^{n+1} x^{n-1} \ dx$$

$$ = \int_{0}^{1} \sum_{n=0}^{\infty} (-1)^{n+2} x^{n} = \int_{0}^{1} \sum_{n=0}^{\infty} (-1)^{n} x^{n} \ dx = \int_{0}^{1}\frac{1}{1+x} \ dx = \ln 2$$
Thanks a lot Random Variable! :)
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
For alternating sums , it's always helpful to look at even and odd terms


\(\displaystyle \sum_{n\geq 0}\frac{(-1)^n}{2n+1}=\sum_{n\geq 0}\frac{1}{4n+1}-\frac{1}{4n+3} \)

where we have for even terms

\(\displaystyle \sum_{n\geq 0}\frac{(-1)^{2n}}{4n+1}=\sum_{n\geq 0}\frac{1}{4n+1}\)

and odd terms

\(\displaystyle \sum_{n\geq 0}\frac{(-1)^{2n+1}}{2(2n+1)+1}=-\sum_{n\geq 0}\frac{1}{4n+3}\)