Limit of Sigma Notation

[Hckyplayer8]

I have no idea. There is an example in the book but it just shows you how to solve it. Not how to understand what this problem is saying.

At first glance...without any experience...I would interpret this as saying, what happens as the upper limit increases without bound. So if k is always one and N can increase towards infinity, then the ratio would yield a very small number which means the limit would be zero.

 

[lev888]

View attachment 14427

I have no idea. There is an example in the book but it just shows you how to solve it. Not how to understand what this problem is saying.

At first glance...without any experience...I would interpret this as saying, what happens as the upper limit increases without bound. So if k is always one and N can increase towards infinity, then the ratio would yield a very small number which means the limit would be zero.

k is not always 1. It iterates from 1 to N. Look up sigma notation.

 

[lev888]

This is a good example:

You have a sequence of sums. Your goal is to get rid of sums and make it a regular sequence, then figure out the limit.

 

[Hckyplayer8]

k is not always 1. It iterates from 1 to N. Look up sigma notation.

So in a sense, N is just a k in disguise?

The two variables are what is throwing me off. I understand k is my lower limit and N is my upper limit but don't really get what it is trying to say by investigating the ratio of the lower limit times two and the upper limit to the second power.

 

[Hckyplayer8]

This is a good example:

You have a sequence of sums. Your goal is to get rid of sums and make it a regular sequence, then figure out the limit.

Thanks! I will have a look.

 

[lev888]

So in a sense, N is just a k in disguise?

The two variables are what is throwing me off. I understand k is my lower limit and N is my upper limit but don't really get what it is trying to say by investigating the ratio of the lower limit times two and the upper limit to the second power.

Let's make sure you understand the notation.
[MATH]\sum_{k=1}^4k [/MATH] = ?
[MATH]\sum_{k=1}^42 [/MATH] = ?
[MATH]\sum_{k=1}^4M [/MATH] = ?

 

[Steven G]

View attachment 14427

I have no idea. There is an example in the book but it just shows you how to solve it. Not how to understand what this problem is saying.

At first glance...without any experience...I would interpret this as saying, what happens as the upper limit increases without bound. So if k is always one and N can increase towards infinity, then the ratio would yield a very small number which means the limit would be zero.

Forget the limit for now. First k=1 so we get 2*1/N^2. Then k = 2 and we get 2*2/N^2. Then k= 3 and we get 2*3/N^2 ... Then k=N, so we get 2*N/N^2
Now we add up all those terms but first notice that they all 2/N^2 in common so we factor it out.
The sum for be (2/N^2)(1+2+3+4 + ... + N) = (2/N^2)(N(N+1)/2) = (N+1)/N
Can you continue from here?


Alternate method. From the very beginning factor out everything that has no k in it. Try it that way and see what you get.


In the end ONLY k changes and you keep adding until k is N.

 

[Steven G]

So in a sense, N is just a k in disguise?

k goes from 1 to 2 to 3 to 4 to ... to N.

 

[pka]

View attachment 14427

\(\displaystyle \sum\limits_{k = 1}^N {\frac{{2k}}{{{N^2}}}} = \frac{2}{{{N^2}}}\sum\limits_{k = 1}^N {k = } \frac{2}{{{N^2}}}\left( {\frac{{N(N + 1)}}{2}} \right) = 1 + \frac{1}{N}\)
Now let \(\displaystyle N\to\infty\). What do you get?

 

[Hckyplayer8]

Let's make sure you understand the notation.
[MATH]\sum_{k=1}^4k [/MATH] = 10
[MATH]\sum_{k=1}^42 [/MATH] = 8
[MATH]\sum_{k=1}^4M [/MATH] =

The last one with the variable I'm not too sure about. I assume you just sum the variable for 4M.

 

[lev888]

Let's make sure you understand the notation.
[MATH]\sum_{k=1}^4k [/MATH] = ?
[MATH]\sum_{k=1}^42 [/MATH] = ?
[MATH]\sum_{k=1}^4M [/MATH] = ?

The last one with the variable I'm not too sure about. I assume you just sum the variable for 4M.

Good.
Can you simplify this?
[MATH]\sum_{k=1}^nk [/MATH]

 

[Hckyplayer8]

Forget the limit for now. First k=1 so we get 2*1/N^2. Then k = 2 and we get 2*2/N^2. Then k= 3 and we get 2*3/N^2 ... Then k=N, so we get 2*N/N^2
Now we add up all those terms but first notice that they all 2/N^2 in common so we factor it out.
The sum for be (2/N^2)(1+2+3+4 + ... + N) = (2/N^2)(N(N+1)/2) = (N+1)/N
Can you continue from here?


Alternate method. From the very beginning factor out everything that has no k in it. Try it that way and see what you get.


In the end ONLY k changes and you keep adding until k is N.

\(\displaystyle \sum\limits_{k = 1}^N {\frac{{2k}}{{{N^2}}}} = \frac{2}{{{N^2}}}\sum\limits_{k = 1}^N {k = } \frac{2}{{{N^2}}}\left( {\frac{{N(N + 1)}}{2}} \right) = 1 + \frac{1}{N}\)
Now let \(\displaystyle N\to\infty\). What do you get?

Thank you both. Once we get the problem down to 1 + 1 / pos infinity it's easy enough to see that the fraction is approaching zero and the ultimate answer will be 1.

I'm following on the concept of the problem is saying the summation of 2*1/n^2 + 2*2/n^2...and so on and so forth.

Its after that where I may need the "slow motion" math.

"The sum for be (2/N^2)(1+2+3+4 + ... + N) = (2/N^2)(N(N+1)/2) = (N+1)/N" this part. Idk if I was doing something wrong but I tried punching the problem into symbolab and it did not like it at all.

 

[Hckyplayer8]

Good.
Can you simplify this?
[MATH]\sum_{k=1}^nk [/MATH]

Can I or is it possible?

To me, it looks like it is in lowest terms.

So I'm thinking the answer to your question is a no.

 

[lev888]

Can I or is it possible?

To me, it looks like it is in lowest terms.

So I'm thinking the answer to your question is a no.

Do you understand post #9? Do you see what happened to this sum there? Maybe simplify was not the right word, but that's what I was looking for.

 

[Steven G]

You really have to be able to handle [MATH]\sum_{k=1}^nk [/MATH]Yes, It can be done.
Don't take this badly but one knows nothing about sigma notation if they can't do this problem. Understand that no one is born knowing this so I an not trying to say any about your skills (at some point I too did not know sigma notation) but do admire that you keep trying and come here to ask questions when you are stuck.

Is it that n in the upper limit that bothers you? The notation says that k goes from 1 to n. Now you need to decide what we are summing up. It happens to be k. So 1st write what ever k equals when k =1 , then when k=2, ..., then when k=n. Now add up this terms or at least list the sum.

Can you please try to do the following ones? If you do not know how to do them please start a new thread for each of them. These are basic ones and if you can do these then you should be able to do any sigma notation problems

1)[MATH]\sum_{k=1}^n1 [/MATH]2)[MATH]\sum_{k=1}^5(k^2 +1) [/MATH]3)[MATH]\sum_{k=1}^4(3k+ 1/k) [/MATH]4)[MATH]\sum_{k=1}^p(k+3n) [/MATH]

 

[pka]

If you doubt my post, perhaps you will follow this?

 

[Hckyplayer8]

If you doubt my post, perhaps you will follow this?

I never said I did. Your post was great. Sometimes I need this subject spelled out one letter at a time which is why I mentioned symbolab.

 

[Hckyplayer8]

You really have to be able to handle [MATH]\sum_{k=1}^nk [/MATH]Yes, It can be done.
Don't take this badly but one knows nothing about sigma notation if they can't do this problem. Understand that no one is born knowing this so I an not trying to say any about your skills (at some point I too did not know sigma notation) but do admire that you keep trying and come here to ask questions when you are stuck.

Is it that n in the upper limit that bothers you? The notation says that k goes from 1 to n. Now you need to decide what we are summing up. It happens to be k. So 1st write what ever k equals when k =1 , then when k=2, ..., then when k=n. Now add up this terms or at least list the sum.

Can you please try to do the following ones? If you do not know how to do them please start a new thread for each of them. These are basic ones and if you can do these then you should be able to do any sigma notation problems

1)[MATH]\sum_{k=1}^n1 [/MATH] You start at 1 and then add infinity by 1.
2)[MATH]\sum_{k=1}^5(k^2 +1) [/MATH]3)[MATH]\sum_{k=1}^4(3k+ 1/k) [/MATH]4)[MATH]\sum_{k=1}^p(k+3n) [/MATH]

1)[MATH]\sum_{k=1}^n1 [/MATH] You start at 1 and then add to infinity by 1. [1+2+3+4...n] Which will be infinity see there is no upper bound.

Is this what I'm doing in Lev888's problem? Since there is no upper bound, the expression just goes to positive infinity?

2)[MATH]\sum_{k=1}^5(k^2 +1) [/MATH] Plug each value of the series of 1-5 in and then sum. I got 60.
3)[MATH]\sum_{k=1}^4(3k+ 1/k) [/MATH] Same deal with this one, except the series is till 4. I got 32.5
4)[MATH]\sum_{k=1}^p(k+3n) [/MATH] This one is harder since there are so many variables. I know how to correlate the k but not the p and n.

 

[pka]

4)[MATH]\sum_{k=1}^p(k+3n) [/MATH] This one is harder since there are so many variables. I know how to correlate the k but not the p and n.

\(\displaystyle \sum\limits_{k = 1}^P {(k + 3n)} = \sum\limits_{k = 1}^P {(k)} + 3\sum\limits_{k = 1}^P {(n)} = \frac{{P(P + 1)}}{2} + 3(P \cdot n)\)
Now you need to post an explanation of that answer.

 

[pka]

2)[MATH]\sum_{k=1}^5(k^2 +1) [/MATH] Plug each value of the series of 1-5 in and then sum. I got 60.
3)[MATH]\sum_{k=1}^4(3k+ 1/k) [/MATH] Same deal with this one, except the series is till 4. I got 32.5
4)[MATH]\sum_{k=1}^p(k+3n) [/MATH] This one is harder since there are so many variables. I know how to correlate the k but not the p and n.

Why don't you learn to use this website to check your work?