Simplifying summations

nightcrrawlerr

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

nightcrrawlerr

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

c.)

[tex] \sum^n _{k=1} \frac{-1}{k(k + 1)} [/tex]


[tex]= \sum^n _{k=1} (\frac{-1}{k} + \frac {1}{k + 1}) [/tex]

[tex]= (\frac {-1}{1} + \frac {1}{2}) + (\frac {-1}{2} + \frac {1}{3}) + ... + (\frac{-1}{n} + \frac {1}{n + 1}) [/tex]



[tex]= {-1} + \frac{1}{n + 1} [/tex]

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

cristo

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

nightcrrawlerr

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

Tom Mattson suggested this:

[tex]\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}[/tex]


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

thanks for helping me out.

cristo

Looks good to me

nightcrrawlerr

You mean my answer here is correct?

Kindly help on this also, please.


[tex]\sum^{\infty}_{k=0}\frac{k-1}{2^k}[/tex]

[tex]=\sum^n_{k=1}\frac{k}{2^k}-\sum^n_{k=0}\frac{1}{2^k}[/tex]


what should I do next?

nightcrrawlerr

What will the answer if i combine both equations?

Is it [tex]-3 ^n^+^1[/tex]

Dick

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

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

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:

[tex]\sum^n_{k=0}(\frac{x}{2})^k=
\frac{1-(\frac{x}{2})^{n+1}}{1-(\frac{x}{2})}[/tex]

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

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.

[tex] \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 [/tex] (1)

[tex] \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) [/tex] (2)

On the other hand

[tex] \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 [/tex] (3)

From (2) and (3) it follows that

[tex] p(1)=2 [/tex] (4)

I have used that

[tex] 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. [/tex] (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

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

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!