Converge or Diverge?

[Jason76]

\(\displaystyle \lim n \rightarrow \infty \sum_{n = 1}^{\infty} \dfrac{n^{2} - 1}{n^{2} + n}\)

First use basic convergence test (Does the limit as n approaches infinity equal zero?) If it equals 0, then further tests should be done. Otherwise, it's divergent.

\(\displaystyle \lim n \rightarrow \infty \sum_{n = 1}^{\infty} \dfrac{(\infty)^{2} - 1}{(\infty)^{2} + (\infty)} = ?\) I know it doesn't equal zero, so from the get go, without running more test, we know it's divergent.

 

[stapel]

I will guess that the instructions (not included within the posting) were actually in the subject line. This is often ("usually"?) not the case, so apologies if this is incorrect.

Subject: Converge or Diverge?

\(\displaystyle \displaystyle{\lim_{n \rightarrow \infty} \sum_{n = 1}^{\infty} \dfrac{n^{2} - 1}{n^{2} + n}}\)

First use basic convergence test (Does the limit as n approaches infinity equal zero?)

Does the limit of what equal zero?

\(\displaystyle \displaystyle{\lim_{n \rightarrow \infty} \sum_{n = 1}^{\infty} \dfrac{(\infty)^{2} - 1}{(\infty)^{2} + (\infty)} = ?}\)

How about first considering the terms. If the terms of the summation don't tend toward zero as the index gets arbitrarily large, then the summation of those terms cannot be finite. Applying what you learned back in algebra, to what value to the individual terms tend?

. . . . .\(\displaystyle \dfrac{n^2\, -\, 1}{n^2\, +\, n}\, =\, \dfrac{n^2 \left(1\, -\, \dfrac{1}{n^2}\right)}{n^2 \left(1\, +\, \dfrac{1}{n}\right)}\, =\, \left(\dfrac{n^2}{n^2}\right)\left(\dfrac{1\, -\, \dfrac{1}{n^2}}{1\, +\, \dfrac{1}{n}}\right)\, =\, \dfrac{1\, -\, \dfrac{1}{n^2}}{1\, +\, \dfrac{1}{n}}\)

To what value does this tend?

 

[Jason76]

Actually the test on this one, comes out to be indeterminate over infinity. So what to do in this case? If this means indeterminate, then you have to divide each term by the highest term in the denominator.

 

[stapel]

Actually the test on this one, comes out to be indeterminate over infinity.

What "test"? "Comes out" how? What do you mean, mathematically, by "indeterminate over infinity"?

What, exactly, are you trying to do here? Are you trying to determine if an infinity series converges? If so, why have you not first determined whether the terms of the series converge to zero (as was explained to you -- and was nearly completed for you -- previously)? Or is this another thread in which you are not reading replies...?

 

[Jason76]

Sorry to hold people up. Will reply in a few hours, with a more clear explanation.

 

[Deleted member 4993]

\(\displaystyle \lim n \rightarrow \infty \sum_{n = 1}^{\infty} \dfrac{n^{2} - 1}{n^{2} + n}\)

First use basic convergence test (Does the limit as n approaches infinity equal zero?) If it equals 0, then further tests should be done. Otherwise, it's divergent.

\(\displaystyle \lim n \rightarrow \infty \sum_{n = 1}^{\infty} \dfrac{(\infty)^{2} - 1}{(\infty)^{2} + (\infty)} = ?\) I know it doesn't equal zero, so from the get go, without running more test, we know it's divergent.

Another way:

\(\displaystyle \displaystyle{\lim_{n \rightarrow \infty} \sum_{n = 1}^{\infty} \dfrac{n^{2} - 1}{n^{2} + n}}\)

\(\displaystyle = \displaystyle{\lim_{n \rightarrow \infty} \sum_{n = 1}^{\infty} \dfrac{(n+1)(n - 1)}{n(n+1)}}\)

\(\displaystyle = \displaystyle{\lim_{n \rightarrow \infty} \sum_{n = 1}^{\infty} \dfrac{(n - 1)}{n}}\)

\(\displaystyle = \displaystyle{\lim_{n \rightarrow \infty} \sum_{n = 1}^{\infty}\left [1 - \dfrac{1}{n}\right ]}\)

\(\displaystyle = \displaystyle{\lim_{n \rightarrow \infty} \sum_{n = 1}^{\infty}1}\)

= ??

 

[Jason76]

Another way:

\(\displaystyle \displaystyle{\lim_{n \rightarrow \infty} \sum_{n = 1}^{\infty} \dfrac{n^{2} - 1}{n^{2} + n}}\)

\(\displaystyle = \displaystyle{\lim_{n \rightarrow \infty} \sum_{n = 1}^{\infty} \dfrac{(n+1)(n - 1)}{n(n+1)}}\)

\(\displaystyle = \displaystyle{\lim_{n \rightarrow \infty} \sum_{n = 1}^{\infty} \dfrac{(n - 1)}{n}}\)

\(\displaystyle = \displaystyle{\lim_{n \rightarrow \infty} \sum_{n = 1}^{\infty}\left [1 - \dfrac{1}{n}\right ]}\)

\(\displaystyle = \displaystyle{\lim_{n \rightarrow \infty} \sum_{n = 1}^{\infty}1}\)

= ??

The limit exists, but what does that mean in this case.?

 

[Deleted member 4993]

The limit exists, but what does that mean in this case.?

What does:

The limit of the sum (series) approaches infinity as n approaches infinity

mean to you?