How can I simplify this summation?

[nightcrrawlerr]

The first term in a) is the only part of this that makes any sense. To make any further progress you are going to have to learn how to sum a geometric series \sum^n _{k=0} (x^k). Could you look that up and post the answer and then apply it to a) to get a correct solution? Once you do that, this thread has all of the hints you need to do b). I think c) may be harder - but let's get a) and b) out of the way anyway. Can you also double check which (if any) of the limits is infinite? I rather hope it's c). And is the (-1) in c) supposed to be (-1)^k?

Based on the equation provided in question c. This is the correct equation.

\sum^n _{k=1} {\frac {^-^1} _{k(k + 1)}}.

Anyway I will verify this also...

again thank you.

 

[berkeman]

Hey, do y'all want me to move this to a General Math forum? I probably made a mistake putting it here. If it would help at all for this to be in a General Math forum, please let me know. Thanks.

 

[nightcrrawlerr]

Hey, do y'all want me to move this to a General Math forum? I probably made a mistake putting it here. If it would help at all for this to be in a General Math forum, please let me know. Thanks.

Ok sure you will. No problem for me.

thanks

 

[berkeman]

Thread moved to the general calculus forum. This is more general than the original impression that I had. Sorry about that. These are hard.

 

[nightcrrawlerr]

Guru, can i start posting my ans. on item c?

 

[nightcrrawlerr]

Dick can i post here my answer for c.)?

c.)

\sum^n _{k=1} \frac{-1}{k(k + 1)}


= \sum^n _{k=1} (\frac{-1}{k} + \frac {1}{k + 1})

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



= {-1} + \frac{1}{n + 1}

cristo, am i doing it right? thanks for your help by the way.

 

[cristo]

Dick can i post here my answer for c.)?

c.)

\sum^n _{k=1} \frac{-1}{k(k + 1)}


\sum^n _{k=1} (\frac{k}{k} - \frac {-1}{k + 1})

Your last line is incorrect. It should read \sum^n _{k=1} (\frac{-1}{k} + \frac {1}{k + 1})

 

[nightcrrawlerr]

Yep, I agree.



I'm not so sure that will help. The summations aren't from 1 to infinity (as they are in a power series), they're from 1 to n.

For b), you could start by splitting it up into 2 sums:

\sum^n_{k=0}\frac{k-1}{2^k}=\sum^n_{k=1}\frac{k}{2^k}-\sum^n_{k=0}\frac{1}{2^k}

Note that the second sum on the RHS only needs to start at 1, not 0. As for the second sum, it's geometric, which makes the sum easy to find. I don't have any clues for you about the first one on the RHS. Try to work it out, and post what you come up with if you get stuck.

For c), I think the best way to tackle it would be to recognize that the summand is telescoping. You can see that by decomposing it into partial fractions, then writing out the first several terms. It should be easy to simplify it from there.

Ok Dick and Tom Mattson, here i am again. kindly help me here out.

Tom Mattson suggested this:

\sum^n_{k=0}\frac{k-1}{2^k}=\sum^n_{k=1}\frac{k}{2^k}-\sum^n_{k=0}\frac{1}{2^k}


How to solved the first and second equation? is it by geometric series still?

thanks for helping me out.

 

[cristo]

Dick can i post here my answer for c.)?

c.)

\sum^n _{k=1} \frac{-1}{k(k + 1)}


= \sum^n _{k=1} (\frac{-1}{k} + \frac {1}{k + 1}) = (\frac {-1}{1} + \frac {1}{2}) + (\frac {-1}{2} + \frac {1}{3}) + ... + (\frac{-1}{n} + \frac {1}{n + 1})


= {-1} + \frac{1}{n + 1}

cristo, am i doing it right? thanks for your help by the way.

Looks good to me

 

[nightcrrawlerr]

Looks good to me

You mean my answer here is correct?

Kindly help on this also, please.


\sum^{\infty}_{k=0}\frac{k-1}{2^k}

=\sum^n_{k=1}\frac{k}{2^k}-\sum^n_{k=0}\frac{1}{2^k}


what should I do next?

 

[nightcrrawlerr]

That looks good. Now that you've got part a) look at part b). There's a pretty strong hint back in the thread. And as Tom suggested so long ago, I think c) is best treated as telescoping series.

What will the answer if i combine both equations?

Is it -3 ^n^+^1

 

[Dick]

What will the answer if i combine both equations?

Is it -3 ^n^+^1

Uh. That's "-3 ^n^+^1" inside the tex brackets. ? Don't combine the two parts. They are too different. Also when you 'combine' two terms you seem to cancel a lot of stuff out at random. I think you need to 1) practice your algebra and 2) as an immediate strategy, avoid doing any that you don't have to. Are you sure this problem set isn't from a level you just aren't ready for yet? Some pretty experienced people here have a hard time with b).

 

[Dick]

I have another suggestion. Take these series and work out the cases n=1,2,3, say (ie get numbers). Then when you get a formula for the answer try putting in n=1,2,3 and see if you get the same thing. It should help you to catch when you are going wrong. Then you can try and figure out why.

 

[slider142]

To those who reply, thanks a lot!

Kindly post also some of the books or reference where I can fully understand this.

Again thanks a lot.

Any basic introductory calculus text should show some tips on series, but one very good inexpensive text that is all about series is "Theory and Application of Infinite Series" by Knopp, available from Dover publishing.

 

[nightcrrawlerr]

Uh. That's "-3 ^n^+^1" inside the tex brackets. ? Don't combine the two parts. They are too different. Also when you 'combine' two terms you seem to cancel a lot of stuff out at random. I think you need to 1) practice your algebra and 2) as an immediate strategy, avoid doing any that you don't have to. Are you sure this problem set isn't from a level you just aren't ready for yet? Some pretty experienced people here have a hard time with b).

I'm just new about summation. They just give us this assignment.

Kindly help me how can I get the correct answer on this. I only have 2 days to finish this up.

So please...

 

[Dick]

You should have everything but the first term in b). That is the hard one. Take this:

\sum^n_{k=0}(\frac{x}{2})^k=<br /> \frac{1-(\frac{x}{2})^{n+1}}{1-(\frac{x}{2})}

It's a special case of the geometric series sum, right? Differentiate both sides and put x=1. This has been posted before. If these instructions don't mean anything to you then I think the series in b) was either assigned by mistake, or you are expected to look it up in a table.

 

[nightcrrawlerr]

You should have everything but the first term in b). That is the hard one. Take this:

= \sum^n_{k=0}(\frac{x}{2})^k=<br /> \frac{1-(\frac{x}{2})^{n+1}}{1-(\frac{x}{2})}

It's a special case of the geometric series sum, right? Differentiate both sides and put x=1. This has been posted before. If these instructions don't mean anything to you then I think the series in b) was either assigned by mistake, or you are expected to look it up in a table.

So dick pls. show how can i start and finish this up pls?

 

[Dick]

Look, nightcrrawlerr. Could you pls. help me? I am not supposed to, nor do I want to, do your homework for you. What I've already posted is really close to doing that and I'm not going any farther. You have an expression and you have explicit instructions for a calculation that will complete it. If you don't know what a derivative is, then this problem was assigned by mistake and I would complain to your instructor.

 

[Gib Z]

If you just want the answer, you should probably try looking it up.

 

[dextercioby]

This is a neat problem and it takes some imagination to solve it. Lemme have a try.

\sum_{k=0}^{\infty} \frac{k-1}{2^{k}}=\left(\sum_{k=1}^{\infty} \frac{k-1}{2^{k}}\right)+\left\frac{k-1}{2^{k}}\right|_{k=0}=\left(\sum_{k=1}^{\infty} \frac{k-1}{2^{k}}\right)-1 (1)

\sum_{k=1}^{\infty} \frac{k-1}{2^{k-1}}=\sum_{\tilde{k}=0}^{\infty} \frac{\tilde{k}}{2^{\tilde{k}}}=\sum_{\tilde{k}=1}^{\infty} \frac{\tilde{k}}{2^{\tilde{k}}}=\sum_{k=1}^{\infty} \frac{k}{2^{k}}\equiv p(1) (2)

On the other hand

\sum_{k=1}^{\infty} \frac{k-1}{2^{k-1}}=2\sum_{k=1}^{\infty} \frac{k-1}{2^{k}}=2\left(\sum_{k=1}^{\infty}\frac{k}{2^{k}}\right)-2\left(\sum_{k=1}^{\infty}\frac{1}{2^{k}} \right)=2p(1)-2 (3)

From (2) and (3) it follows that

p(1)=2 (4)

I have used that

2\sum_{k=1}^{\infty}\frac{1}{2^{k}} =2 \left [\left(\sum_{k=0}^{\infty}\frac{1}{2^{k}}\right)-\left\frac{1}{2^{k}}\right|_{k=0} \right]=2(2-1)=2. (5)


Therefore the initial sum is 2-1=1.

BTW, i just realized that i posted a complete solution. Actually the original poster is welcomed to think of some other method to find it. I hope my solution is not the only one possible.

 

[nightcrrawlerr]

Look, nightcrrawlerr. Could you pls. help me? I am not supposed to, nor do I want to, do your homework for you. What I've already posted is really close to doing that and I'm not going any farther. You have an expression and you have explicit instructions for a calculation that will complete it. If you don't know what a derivative is, then this problem was assigned by mistake and I would complain to your instructor.

Dick, I do appreciate your response to my post. Your answers and explanations.

You probably correct friend. Honestly I took this course twice already at first I drop this because I know I can't make it. For honest reason, I took this for the second time. Why? Because its part of my curriculum and I can't scape it.

Can I took "Algorithm and analysis" by just studying Discrete Math?

Since I don't really have any ideas of what a "Derivative" is, I have to read a lot of books just to understand what is this all about, right? But this is a study of algorithm and analysis and I supposed somebody could explain it well and help me in my comprehension so as through with this derivative.

With this, I understand I couldn't make it. I concede. This is not my cup of tea...

Again to those who help me out in this, thanks a lot.

To Dick, Mattson, slider142, cristo, dextercioby and to those who viewed this thread others who participated in here... thanks and more power.

May the grace our Lord Jesus be with you always.


Adios!

 

[Dick]

nightcrrawlerr,

I'm really sorry this has been so hard for you. I thought calculus at at least a basic level was required for a course like this. If not, then they are making a mistake including problems like this. I wish you luck in all your studies!

Dick

 

[Dick]

Therefore the initial sum is 2-1=1.

I'm SURE there is another way to do this. Because the infinite version of this series sums to 0.

 

[Dick]

Therefore the initial sum is 2-1=1.

Should read 2-2=0. That's the only problem. So there IS a way to do this without derivatives.

 

[Gib Z]

2\sum_{k=1}^{\infty}\frac{1}{2^{k}} =2 \left [\left(\sum_{k=0}^{\infty}\frac{1}{2^{k}}\right)-\left\frac{1}{2^{k}}\right|_{k=0} \right]=2(2-1)=2.

Quote dex: I hope this isn't the only way of solving this
I think u ment the entire solution, but for that summation here's a nice visual:

1x1 unit square. First term of series, 1/2, color in half the square. 2nd term, color in 1/4 of the remaining unshaded part. Then next term, 1/8, do the same. You'll see its going to fill it up. :D