# Simplifying summations

## Main Question or Discussion Point

1. Homework Statement

Simplify the summation statement.

2. Homework Equations

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

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

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

3. The Attempt at a Solution

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

thank you.

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1. Homework Statement

Simplify the summation statement.

2. Homework Equations

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

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

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

3. The Attempt at a Solution

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

thank you.

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

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

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

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Dick
Homework Helper
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
Mentor
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
Homework Helper
Clearly. The intent was so obvious I missed the misplaced = sign.

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

berkeman
Mentor
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.

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Dick
Homework Helper
Crabby retort expunged.

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berkeman
Mentor
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.

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

Tom Mattson
Staff Emeritus
Gold Member
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
Homework Helper
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.

Tom Mattson
Staff Emeritus
Gold Member
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"?

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.

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

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

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

$$= 1 - 1 / n$$

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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?

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

Thanks.

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

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

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

Tom Mattson
Staff Emeritus
Gold Member
Still I couldn't figure it out.

By the way what's RHS?
Hehe, sorry.

RHS=Right Hand Side
LHS=Left Hand Side

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 recomend 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".

Tom Mattson
Staff Emeritus
Gold Member
$$\sum^n_{k=0}(\frac{x}{2})^k= \frac{1-(\frac{x}{2})^{n+1}}{1-(\frac{x}{2})}$$

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

Tom Mattson
Staff Emeritus
Gold Member
I call it a sum. :tongue2:

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

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
Homework Helper
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?

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.

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