Proof for Convergent of Series With Seq. Similar to 1/n

In summary, the given series is divergent because it can be lower bounded by the harmonic series, which is known to be divergent.
  • #1
The-Mad-Lisper
12
1

Homework Statement


[itex]\sum\limits_{n=1}^{\infty}\frac{n-1}{(n+2)(n+3)}[/itex]

Homework Equations


[itex]S=\sum\limits_{n=1}^{\infty}a_n[/itex] (1)
[itex]\lim\limits_{n\rightarrow\infty}\frac{a_{n+1}}{a_n}\gt 1\rightarrow S\ is\ divergent[/itex] (2)
[itex]\lim\limits_{n\rightarrow\infty}\frac{a_{n+1}}{a_n}\lt 1\rightarrow S\ is\ convergent[/itex] (3)
[itex]\sum\limits_{n=1}^{\infty}\mid a_n\mid\ is\ convergent\ \land\ \mid b_n \mid\ \leq\ \mid a_n \mid\ for\ every\ n\rightarrow \sum\limits_{n=1}^{\infty}\mid b_n\mid\ is\ convergent[/itex] (4)

The Attempt at a Solution


I suspect that the given series is divergent, thus the series would not be convergent. Thus I take the contrapositive of statement 4 which comes out to be [itex]\sum\limits_{n=1}^{\infty}\mid b_n\mid\ is\ divergent\ \land\ \mid b_n \mid\ \leq\ \mid a_n \mid\ for\ every\ n\rightarrow \sum\limits_{n=1}^{\infty}\mid a_n\mid\ is\ divergent[/itex].
Let [itex]a_n=\frac{n-1}{(n+2)(n+3)}[/itex] and [itex]b_n = \frac{1}{n}[/itex]. The statement [itex]\mid b_n \mid\ \leq\ \mid a_n[/itex] is never true in the domain [itex][1,\infty)[/itex], so we cannot say that [itex]a_n[/itex] is divergent. Any suggestions?
 
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  • #2
The-Mad-Lisper said:

Homework Statement


[itex]\sum\limits_{n=1}^{\infty}\frac{n-1}{(n+2)(n+3)}[/itex]

Homework Equations


[itex]S=\sum\limits_{n=1}^{\infty}a_n[/itex] (1)
[itex]\lim\limits_{n\rightarrow\infty}\frac{a_{n+1}}{a_n}\gt 1\rightarrow S\ is\ divergent[/itex] (2)
[itex]\lim\limits_{n\rightarrow\infty}\frac{a_{n+1}}{a_n}\lt 1\rightarrow S\ is\ convergent[/itex] (3)
[itex]\sum\limits_{n=1}^{\infty}\mid a_n\mid\ is\ convergent\ \land\ \mid b_n \mid\ \leq\ \mid a_n \mid\ for\ every\ n\rightarrow \sum\limits_{n=1}^{\infty}\mid b_n\mid\ is\ convergent[/itex] (4)

The Attempt at a Solution


I suspect that the given series is divergent, thus the series would not be convergent. Thus I take the contrapositive of statement 4 which comes out to be [itex]\sum\limits_{n=1}^{\infty}\mid b_n\mid\ is\ divergent\ \land\ \mid b_n \mid\ \leq\ \mid a_n \mid\ for\ every\ n\rightarrow \sum\limits_{n=1}^{\infty}\mid a_n\mid\ is\ divergent[/itex].
Let [itex]a_n=\frac{n-1}{(n+2)(n+3)}[/itex] and [itex]b_n = \frac{1}{n}[/itex]. The statement [itex]\mid b_n \mid\ \leq\ \mid a_n[/itex] is never true in the domain [itex][1,\infty)[/itex], so we cannot say that [itex]a_n[/itex] is divergent. Any suggestions?
That's a good idea, comparing with the harmonic series ##\sum \frac{1}{n}##.

Hint: ##\frac{n-1}{n+2}## is close to 1 for large n.
 
  • #3
@The-Mad-Lisper Have you studied the limit comparison test yet?
 
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  • #4
LCKurtz said:
@The-Mad-Lisper Have you studied the limit comparison test yet?
Only after the fact, I managed to modify Nicole Oresme's proof of the divergent nature of a harmonic series.
 
  • #5
The-Mad-Lisper said:

Homework Statement


[itex]\sum\limits_{n=1}^{\infty}\frac{n-1}{(n+2)(n+3)}[/itex]

Homework Equations


[itex]S=\sum\limits_{n=1}^{\infty}a_n[/itex] (1)
[itex]\lim\limits_{n\rightarrow\infty}\frac{a_{n+1}}{a_n}\gt 1\rightarrow S\ is\ divergent[/itex] (2)
[itex]\lim\limits_{n\rightarrow\infty}\frac{a_{n+1}}{a_n}\lt 1\rightarrow S\ is\ convergent[/itex] (3)
[itex]\sum\limits_{n=1}^{\infty}\mid a_n\mid\ is\ convergent\ \land\ \mid b_n \mid\ \leq\ \mid a_n \mid\ for\ every\ n\rightarrow \sum\limits_{n=1}^{\infty}\mid b_n\mid\ is\ convergent[/itex] (4)

The Attempt at a Solution


I suspect that the given series is divergent, thus the series would not be convergent. Thus I take the contrapositive of statement 4 which comes out to be [itex]\sum\limits_{n=1}^{\infty}\mid b_n\mid\ is\ divergent\ \land\ \mid b_n \mid\ \leq\ \mid a_n \mid\ for\ every\ n\rightarrow \sum\limits_{n=1}^{\infty}\mid a_n\mid\ is\ divergent[/itex].
Let [itex]a_n=\frac{n-1}{(n+2)(n+3)}[/itex] and [itex]b_n = \frac{1}{n}[/itex]. The statement [itex]\mid b_n \mid\ \leq\ \mid a_n[/itex] is never true in the domain [itex][1,\infty)[/itex], so we cannot say that [itex]a_n[/itex] is divergent. Any suggestions?

The terms of your sum are all positive, so it is enough to give a lower bound series that diverges.

For ##n > 6## we have ##n-1 > \frac{1}{2}(n+3)## and ##(n+2)(n+3) < (n+3)^2##, so
[tex] \sum_{n=6}^{\infty} \frac{n-1}{(n+2)(n+3)} > \sum_{n=6}^{\infty} \frac{1}{2} \frac{n+3}{(n+3)^2}
= \frac{1}{2} \sum_{k=9}^{\infty} \frac{1}{k} [/tex]
 

FAQ: Proof for Convergent of Series With Seq. Similar to 1/n

1. What is a series with a sequence similar to 1/n?

A series with a sequence similar to 1/n is a series where the terms follow the pattern of 1/n, where n is a positive integer. For example, the series 1/1, 1/2, 1/3, 1/4, ... is a series with a sequence similar to 1/n.

2. What does it mean for a series to converge?

A series converges if the sum of its terms approaches a finite value as the number of terms increases. In other words, as more terms are added to the series, the sum of these terms gets closer and closer to a specific number.

3. How can I prove that a series with a sequence similar to 1/n converges?

To prove that a series with a sequence similar to 1/n converges, you can use the integral test or the comparison test. The integral test compares the series to an improper integral and the comparison test compares the series to another series with known convergence. If the integral or the series used in the comparison test converge, then the original series also converges.

4. Are there any other tests that can be used to prove the convergence of a series with a sequence similar to 1/n?

Yes, there are other tests that can be used to prove the convergence of a series with a sequence similar to 1/n, such as the ratio test and the root test. These tests compare the ratio or the nth root of the terms in the series to a known value to determine convergence.

5. What is the importance of proving the convergence of a series with a sequence similar to 1/n?

Proving the convergence of a series with a sequence similar to 1/n is important because it helps determine the behavior of the series and the sum of its terms. It also allows for the use of the series in various mathematical calculations and applications.

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