How can I simplify this summation?

[nightcrrawlerr]

Homework Statement

Simplify the summation statement.

Homework Equations

a.) \sum^n _{k=0} = \frac{k-1}{2^k}


b.) \sum^{\infty} _{k=0} = (3k - 3^k)


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

The Attempt at a Solution

Please I need your help.

Kindly post here some link of the tutorial for Algorithm analysis.

thank you.

 

[nightcrrawlerr]

Homework Statement

Simplify the summation statement.

Homework Equations

a.) \sum^n _{k=0} = \frac{k-1}{2^k}


b.) \sum^{\infty} _{k=0} = (3k - 3^k)


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

The Attempt at a Solution

Please I need your help.

Kindly post here some link of the tutorial for Algorithm analysis.

thank you.

b.) \sum^{\infty} _{k=0} = (3k - 3^k)




= \sum^{\infty} _{k=0} 3k - \sum^{\infty} _{k=0} 3^k)



= 3(n(n+1)/2) - c3^n^+^1)

please continue...

 

[Dick]

Note for b) the sum is infinite. This makes it easy. For a) and c) try to think of an easy power series in x that can give you terms like these if you put x=1. Remember you can get factors of k, k-1 etc by differentiating and 1/k,1/(k+1) by integrating your power series. Nuff hints?

 

[berkeman]

What in the world are you folks talking about? Sorry if this a standard form that I'm not familiar with, but what is the object of the sum in each case? In my experience, the summation notation all by itself is meaningless. You need to do a sum of something over the limits. Do you mean this?

a.) \sum^n _{k=0} \frac{k-1}{2^k} = ?

 

[Dick]

Clearly. The intent was so obvious I missed the misplaced = sign.

 

[Dick]

Picky..... I've been alerted that I need at least 10 characters in a response.

 

[berkeman]

Okay, sorry about my flip response. I had problems back in my freshman college year when I didn't understand how important the format of an equation was, and being careful about equal signs helped me a lot to understand what an equation means.


EDIT -- I removed some insulting words from my post -- apologies to Dick.

 

[Dick]

Crabby retort expunged.

 

[berkeman]

You're right, that was not a good thing for me to say. I'll delete it out of my post. I guess it just struck a chord with me -- early in my college work, I had a fundamental lack of understanding of how equations worked, and that slowed my initial progress. I apologize for the chiding question.

 

[Dick]

And I apologize for taking offense where none was meant. I'll try and delete my crabby retort.

 

[quantumdude]

Note for b) the sum is infinite. This makes it easy.

Yep, I agree.

For a) and c) try to think of an easy power series in x that can give you terms like these if you put x=1. Remember you can get factors of k, k-1 etc by differentiating and 1/k,1/(k+1) by integrating your power series. Nuff hints?

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.

 

[Dick]

A FINITE geometric series can be summed in closed form as well. And differentiated, and integrated. Consider a finite sum of terms like (x/2)^k. Differentiate it, set x=1 and compare with a). For c) consider integrating a similar finite sum.

 

[quantumdude]

A FINITE geometric series can be summed in closed form as well.

I know that. In fact, I said it!

What I don't see is the connection between power series and finite sums. Could you please explain what you meant by "For a) and c) try to think of an easy power series in x that can give you terms like these if you put x=1"?

 

[nightcrrawlerr]

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.

 

[nightcrrawlerr]

What in the world are you folks talking about? Sorry if this a standard form that I'm not familiar with, but what is the object of the sum in each case? In my experience, the summation notation all by itself is meaningless. You need to do a sum of something over the limits. Do you mean this?

a.) \sum^n _{k=0} \frac{k-1}{2^k} = ?

Sorry berkeman, I shouldn't put the equal sign there. Thank you.

Now I understand, no body answer this post because of that. Anyway its my fault, I hope you understand.

Wish to correct that equation here:

a. ) \sum^n _{k=0} (3k - 3^k)

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

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

Kindly comment with my answer.

for a. ) = \sum^n _{k=0} (3k - 3^k)


= 3 \sum^n _{k=0} k - \sum^n _{k=0} 3^k

= 3 (n(n + 1)/2) - 3n

= 3 - 3n

= n


for b.) \sum^{\infty} _{k=0} \frac{k-1}{2^k}


i have no answer yet... pls help.


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


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

= 1 - 1 / n

 

[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.

Thanks Guru, great help.

Still I couldn't figure it out. By the way what's RHS?

Need help please.

Could you recommend some reading, books, website links, and etc. to further my study of this, please.

Thanks.

 

[Dick]

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

Right? Assuming I did the tex ok. Differentiate, put x=1. The LHS is the tough part of a). The RHS is the answer.

 

[Dick]

I note in the reposted equations a) and b) have undergone a recombination.

 

[quantumdude]

Still I couldn't figure it out.

By the way what's RHS?

Hehe, sorry.

RHS=Right Hand Side
LHS=Left Hand Side

Need help please.

The only part of this that wasn't answered as of my last post, has now been answered by Dick. Can you be specific about what is still giving you trouble?

Could you recommend some reading, books, website links, and etc. to further my study of this, please.

I am drawing only on my knowledge of what we here in the States call "Calculus II".

 

[quantumdude]

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

Right? Assuming I did the tex ok. Differentiate, put x=1. The LHS is the tough part of a). The RHS is the answer.

OK, I understand you now. The part that confused me was when you said to use a "power series". What you have posted above is not a power series, because it doesn't have infinitely many terms.

Though I suppose you could turn it into a power series by letting the index run to infinity and multiplying the summand by a coefficient a_k that is 1 for 0 \leq k \leq n and 0 for n+1 \leq k \leq \infty. Meh, whatever.

 

[Dick]

Can I call it 'finite power series'? What do you call it?

 

[quantumdude]

I call it a sum.

Edited to add:

Actually, it exactly meets the definition of an n-th degree polynomial (with specific coefficients). So I'd call it a polynomial.

 

[nightcrrawlerr]

Kindly comment with my answer.

Kindly help me out of my answer please. If my answer is not correct kindly help me correct it out by teaching me, please.

Thanks a lot for your great help guys. God bless and more power.

for a. ) = \sum^n _{k=0} (3k - 3^k)


= 3 \sum^n _{k=0} k - \sum^n _{k=0} 3^k

= 3 (n(n + 1)/2) - 3n

= 3 - 3n

= n


for b.) \sum^{\infty} _{k=0} \frac{k-1}{2^k}


i have no answer yet... pls help.


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


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


= 1 - 1 / n

 

[Dick]

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?

 

[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?

Dick, thanks for you immediate answer.

Ok i'll try first your comment on this and i will post it later on.

Lets start first at letter a)

\sum^n _{k=0} (3k - 3^k)

this can be solved by telescoping series right?

does this simplifying make sense sir?

= 3 \sum^n _{k=0} k - \sum^n _{k=0} 3^k

then what should i do after this? Kindly post example problems and further solutions to explain it well, please.

thanks a lot.

 

[Dick]

Dick, thanks for you immediate answer.

Ok i'll try first your comment on this and i will post it later on.

Lets start first at letter a)

\sum^n _{k=0} (3k - 3^k)

this can be solved by telescoping series right?

does this simplifying make sense sir?

= 3 \sum^n _{k=0} k - \sum^n _{k=0} 3^k

No, it's not a telescoping series, but your smplification is fine. You know what to do with the first term, right? Use the same formula you used before. As I keep saying, there is also a formula for the second term. It's called a 'geometric series'. Find out about it!

 

[nightcrrawlerr]

No, it's not a telescoping series, but your smplification is fine. You know what to do with the first term, right? Use the same formula you used before. As I keep saying, there is also a formula for the second term. It's called a 'geometric series'. Find out about it!

for geometric series:
= \sum^n _{k=0} a^k = a + a^1 + a^2 + ... + a^n

for a "not equal" 1

= \sum^n _{k=0} 3^k = 3 + 3^1 + 3^2 + ... + 3^n

multiply both sides with (1-a)

(1 - a)\sum^n _{k=0} 3^k = \sum^n _{k=0} 3^k - \sum^n _{k=0} 3^k+^1

= \sum^n _{k=0} 3^k - \sum^n+^1 _{k=1} 3^k

= 3 + 3^1 + 3^2 + ... + 3^n
- 3^1 - 3^2 - ... - 3^n - 3^n+1

= 3 - 3^n+1
= 3 - 3^n+^1

(1 - a) \sum^n _{k=0} 3^k = 3 - 3^n+^1/ 1 - a

\sum^n _{k=0} 3^k = 3 - 3^n+^1 / 1 - a

for P "not equal" 1

 

[nightcrrawlerr]

No, it's not a telescoping series, but your smplification is fine. You know what to do with the first term, right? Use the same formula you used before. As I keep saying, there is also a formula for the second term. It's called a 'geometric series'. Find out about it!

for geometric series:
= \sum^n _{k=0} a^k = a + a^1 + a^2 + ... + a^n

for a "not equal" 1

= \sum^n _{k=0} 3^k = 3 + 3^1 + 3^2 + ... + 3^n

multiply both sides with (1-a)

(1 - a)\sum^n _{k=0} 3^k = \sum^n _{k=0} 3^k - \sum^n _{k=0} 3^k^+^1

= \sum^n _{k=0} 3^k - \sum ^{^n^+^1} _{k=1} 3^k

= 3 + 3^1 + 3^2 + ... + 3^n

= - 3^1 - 3^2 - ... - 3^n - 3^n^+^1

= 3 - 3^n^+^1


(1 - 3) \sum^n _{k=0} 3^k = 3 - 3^n^+^1/ 1 - 3

\sum^n _{k=0} 3^k = 3 - 3^n^+^1 / 1 - 3


for a "not equal" 1

kindly please comment on this pls.

 

[Dick]

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.

 

[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.

So you mean got the solution correct? how about the first term, how can i combine the answers in one equation?

Kindly help me more, please.

thank you.

 

[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.