Indefinite Series

N

NaN-Gram

New member
Joined
Mar 15, 2020
Messages
37
I have been having some issues with these two problems. I don't know how to properly adapt the "ar" system with these expressions. I could use a hint to help get me on the right track.

problem 59.JPG

problem 73.JPG
 
I did try separating 1/e as r, but I don't know how to proceed on that problem otherwise.
 
As for the dual fraction question, I realize that I can also split it into two separate questions as well. Those would be n/(n+1) and (n+2)/(n+3). Again, I got stuck at that point, and I need help to proceed on both questions.
 
How do you get that solution, though? I don't know how to actually equate that.
 
NaN-Gram said:
How do you get that solution, though? I don't know how to actually equate that.
Click to expand...
I think that it is fair to say that students are just expected to have learned some basic facts about the topics.
For example: If \(|r|<1\) then \(\sum\limits_{n = K}^\infty {A{r^n}}\) is the summation of a geometric sequence in which \(Ar^K\) is the first term and \(r\) is known as the common ratio. Terms may differ from course to course. But the sum equals \(\dfrac{Ar^K}{1-r}\): in words, the sum is the first term divided by one minus the common ratio.
 
NaN-Gram said:
As for the dual fraction question, I realize that I can also split it into two separate questions as well. Those would be n/(n+1) and (n+2)/(n+3). Again, I got stuck at that point, and I need help to proceed on both questions.
Click to expand...
If by "questions" you mean "summations", you can't split them, because neither of them converges, as the terms in each case approach 1, not 0. That also implies that their sum approaches 2, so the whole summation can't converge. I agree with topsquark in questioning it.
 
I'm supposed to show why an expression might diverge, or what the expression converges to.
 
NaN-Gram said:
I'm supposed to show why an expression might diverge, or what the expression converges to.
Click to expand...

It's very important, as mentioned here, that you state the entire problem you are asking about, including its instructions! Now we know.

Have you been told enough now to answer the question, in one case showing what it converges to, and in the other why it doesn't converge?
 
I still don't know how to handle the twin sums. I feel that I need a hint to continue, aside from splitting up the expression.
 
NaN-Gram said:
I still don't know how to handle the twin sums. I feel that I need a hint to continue, aside from splitting up the expression.
Click to expand...
Combine the two fractions:
[math]\dfrac{n}{n + 1} + \dfrac{n + 2}{n + 3} = \dfrac{(n)(n + 3) + (n + 2)(n + 1)}{(n +1)(n + 3)} = \dfrac{n^2 + 3n + n^2 + 3n + 2}{(n + 1)(n + 3)} = \dfrac{2n^2 + 6n + 2}{(n + 1)(n + 3)}[/math]
[math]\sum_{n = 1}^{ \infty } \left ( \dfrac{n}{n + 1} + \dfrac{n + 2}{n + 3} \right ) = \sum_{n = 1}^{ \infty } \dfrac{2n^2 + 6n + 2}{(n + 1)(n + 3)}[/math]
So the term [math]a_n = \dfrac{2n^2 + 6n + 2}{(n + 1)(n + 3)}[/math]. There are a bunch of tests for convergence you can use from this point on but probably the simplest is to find [math]\lim_{n \to \infty } \dfrac{a_{n + 1}}{a_n}[/math]. Give it a go.

-Dan
 
You must log in or register to reply here.
Top