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]

$$\sum^n_{k=0}(\frac{x}{2})^k= \frac{1-(\frac{x}{2})^{n+1}}{1-(\frac{x}{2})}$$

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]

$$\sum^n_{k=0}(\frac{x}{2})^k= \frac{1-(\frac{x}{2})^{n+1}}{1-(\frac{x}{2})}$$

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. :tongue2:

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?