# Thread: Help with Conv/Div Series

1. ## Help with Conv/Div Series

This problem has to do with Telescoping Sums Method.

Summation 2 / (n^2 + 4n + 3)

The method the professor taught us was to treat this like a Partial Fractions problem, break it up, find A and B and the result from what I get is

lim t-> infinity, summation ((1 / (n+1)) - (1 / (n+3)).

From here he told us to find the first few sequences by plugging in 1, 2, 3, 4, 5, 6 etc until we notice a pattern of terms cancelling out. All up to this part is cake... what i'm having trouble understanding is how to get the very last few terms where the terms will have t in it. How am I supposed to figure this out? Thanks

2. If you expand out as mentioned, you should notice that all cancel except

1/2+1/3=5/6

That is the sum of the series.

3. Originally Posted by galactus
If you expand out as mentioned, you should notice that all cancel except

1/2+1/3=5/6

That is the sum of the series.

Yea I was able to get that but apparently the professor also wants us to show the very last terms of the sequence. So it'll show something like ( 1 / (t+1) - 1 / t) or something like it. This is what i'm having trouble getting.

To make it more clear. When I start the sequence, it'll be (1/2 - 1/4) + (1/3 - 1/5) + (1/4 - 1/6) + (1/5 - 1/7) +......( This part is what I don't know how to get. The last terms where it'll have whatever letter representing infinity).

4. Originally Posted by JonathanEyoon
This problem has to do with Telescoping Sums Method.

Summation 2 / (n^2 + 4n + 3)

The method the professor taught us was to treat this like a Partial Fractions problem, break it up, find A and B and the result from what I get is

lim t-> infinity, summation ((1 / (n+1)) - (1 / (n+3)).

From here he told us to find the first few sequences by plugging in 1, 2, 3, 4, 5, 6 etc until we notice a pattern of terms cancelling out. All up to this part is cake... what i'm having trouble understanding is how to get the very last few terms where the terms will have t in it. How am I supposed to figure this out? Thanks
If you actually want what you said in your title

Therefore covnergent.

You could also say that

Furhtermore

And

Therefore convergent...you could also do the limit comparison test with zeta(2)

5. Don't really know what's going on above and I already know that it'll be convergent since whatever the first few terms that do not cancel out, it's what it usually converges to. But for the sake of writing things in correct format, the professor wants us to know how to find the LAST terms of the sequence and state them when taking the limit.

6. Originally Posted by JonathanEyoon
Don't really know what's going on above and I already know that it'll be convergent since whatever the first few terms that do not cancel out, it's what it usually converges to. But for the sake of writing things in correct format, the professor wants us to know how to find the LAST terms of the sequence and state them when taking the limit.
Ok

So do this! You already decomposed it and found the relevant non-cancelling terms right? Well convert it back!

So the last term would be

7. What do you mean compose it back? Sorry i'm having a really hard time understanding this. In the pdf file i've just put up, #27 is a telescoping sums problem. I'm having trouble understanding how to derive those last terms where it says ....+(1/ (n - 3) - 1 / (n - 1)) + ( 1 / (n - 2) - 1/ n ). And when put into the problem, only a couple of the terms listed above are kept. Why? I'd like to understand how this is being done.

8. Originally Posted by JonathanEyoon
What do you mean compose it back? Sorry i'm having a really hard time understanding this. In the pdf file i've just put up, #27 is a telescoping sums problem. I'm having trouble understanding how to derive those last terms where it says ....+(1/ (n - 3) - 1 / (n - 1)) + ( 1 / (n - 2) - 1/ n ). And when put into the problem, only a couple of the terms listed above are kept. Why? I'd like to understand how this is being done.
Ok imagine this

If we expand this we get

Since it is absolutely convergent we may rearange the terms to get

So we can see that all except the first and last term will get cancelled out, and as I said the last term was above

Apply the same concept here

9. Ok the above example is the same exact problem in my book and the same one my professor used during the lecture. I just have a multitude of questions between this problem and others. Ok for the example you did, the professor's and the book's last term is - 1/(n + 1)

So then take the limit of 1 - 1/(n+1)

why did you leave the problem the way you did instead of this way? What is the difference. Jeez i'm getting more and more confused . Can you show me by listing the last few terms like the pdf file #27 for the original problem I put up on here? Since the first term and the third term doesn't cancel out, I should have 1 / n and something else right?

10. Originally Posted by JonathanEyoon
Ok the above example is the same exact problem in my book and the same one my professor used during the lecture. I just have a multitude of questions between this problem and others. Ok for the example you did, the professor's and the book's last term is - 1/(n + 1)

So then take the limit of 1 - 1/(n+1)

why did you leave the problem the way you did instead of this way? What is the difference. Jeez i'm getting more and more confused . Can you show me by listing the last few terms like the pdf file #27 for the original problem I put up on here? Since the first term and the third term doesn't cancel out, I should have 1 / n and something else right?
Me and your professor are both correct. Would you like me to do a different example?

11. yes please. Can you do the problem that I posted above?

2 / (n^2 + 4n + 3)

12. In the same fashion, split the original fraction into two ratios and find the telescoping series.