Finding a Convergent Sequence with a Limit of 1

In summary, the ratio test is a way to determine if an infinite series converges or diverges by looking at the limit of the ratio of consecutive terms. If the limit is less than 1, the series converges absolutely. If the limit is greater than 1, the series diverges. If the limit is exactly 1, the test is inconclusive and further analysis is needed to determine convergence or divergence. This can be seen in the example given, where the sequence satisfies the limit condition but the series can still converge or diverge.
  • #1
DPMachine
26
0

Homework Statement



Give an example of a sequence [tex](a_n)[/tex] so that [tex]lim_{n\rightarrow\infty} \left|a_{n+1}/a_{n}\right| =1[/tex] and [tex]\sum^{\infty}_{n=1} a_{n}[/tex] converges

Homework Equations



(Maybe relevant, maybe not)
Theorem which states:

If [tex]\sum^{\infty}_{n=1} a_{n}[/tex] converges, then [tex]lim_{n\rightarrow\infty} a_{n} =0[/tex]

The Attempt at a Solution



I'm having trouble coming up with [tex]\sum^{\infty}_{n=1} a_{n}[/tex] that converges...

Since [tex]lim_{n\rightarrow\infty} a_{n} =0[/tex] doesn't imply the convergence of [tex]\sum^{\infty}_{n=1} a_{n}[/tex] (the theorem only works the other way around), I'm not sure where to start.

Any hint will be appreciated. Thank you.
 
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  • #2
DPMachine said:
I'm having trouble coming up with [tex]\sum^{\infty}_{n=1} a_{n}[/tex] that converges...
You can certainly come up with some converging sum, ignoring the other requirement for a moment?

[tex]a_n=\frac{1}{n}[/tex] satisfies [tex]\left|\frac{a_{n+1}}{a_n}\right|\to 1[/tex], but [tex]\sum^{\infty}_{n=1} a_{n}[/tex] does not converge. Can you modify this example such that it does?
 
  • #3
Landau said:
You can certainly come up with some converging sum, ignoring the other requirement for a moment?

[tex]a_n=\frac{1}{n}[/tex] satisfies [tex]\left|\frac{a_{n+1}}{a_n}\right|\to 1[/tex], but [tex]\sum^{\infty}_{n=1} a_{n}[/tex] does not converge. Can you modify this example such that it does?

Sorry, I don't think my question was clear. I'm having trouble understanding what makes certain partial sums converge/not converge.

So yeah, [tex]a_n=\frac{1}{n}[/tex] does satisfy [tex]\left|\frac{a_{n+1}}{a_n}\right|\to 1[/tex], but why wouldn't [tex]\sum^{\infty}_{n=1} a_{n}[/tex] converge? Isn't [tex]1/n[/tex] approaching zero?
 
  • #4
Last edited:
  • #5
The purpose of this exercise seems to be exploring the edge case of the ratio test. For the terms in an infinite series [itex]\sum a_n[/itex], you look at the limit
[tex]\lim_{n \rightarrow \infty}\left|\frac{a_{n + 1}}{a_n}\right|~=~L[/tex]
If L < 1, the series converges absolutely.
If L > 1, the series diverges.
If L = 1, or no limit exists, the test is inconclusive.

This problem seems to be about that third possibility, where L = 1.
 
  • #6
Mark44 said:
The purpose of this exercise seems to be exploring the edge case of the ratio test. For the terms in an infinite series [itex]\sum a_n[/itex], you look at the limit
[tex]\lim_{n \rightarrow \infty}\left|\frac{a_{n + 1}}{a_n}\right|~=~L[/tex]
If L < 1, the series converges absolutely.
If L > 1, the series diverges.
If L = 1, or no limit exists, the test is inconclusive.

This problem seems to be about that third possibility, where L = 1.

Okay, I think I understand it... so if L=1, it's possible for [tex]\sum a_{n}[/tex] to either converge or diverge.

So, for example, when [tex]a_{n}[/tex] is an alternating harmonic series (from the wiki article above):
1c17b395c60e0e044a7235fc4dcc6915.png


then [tex]\sum a_{n}[/tex] would converge and also have L=1.

On the other hand, if [tex]a_{n}[/tex] is something like [tex]a_{1}=1, a_{n+1}=a_{n}[/tex] so that [tex](a_{n}) = (1, 1, 1, 1, 1, ...)[/tex]

then [tex]\sum a_{n}[/tex] does not converge (since it goes to infinity) but still have L=1.

Is that right?
 
  • #7
Right.
 

1. What is convergence of partial sums?

Convergence of partial sums refers to the behavior of a sequence of partial sums in a series. It describes whether the sequence approaches a specific value as the number of terms increases, or if it diverges (does not approach a specific value).

2. How do you determine if a series has convergent or divergent partial sums?

To determine the convergence or divergence of partial sums, you can use various methods such as the ratio test, the comparison test, or the integral test. These methods involve examining the behavior of the terms in the series and determining if they approach a specific value or not.

3. What is the significance of convergence of partial sums in mathematics?

The convergence of partial sums plays a crucial role in mathematics, particularly in calculus and analysis. It allows us to determine the behavior of a series and make conclusions about its overall convergence or divergence. It is also used in many real-world applications, such as in physics, engineering, and finance.

4. Can a series have both convergent and divergent partial sums?

No, a series can only have either convergent or divergent partial sums. If a series has convergent partial sums, it means that the series overall converges. On the other hand, if a series has divergent partial sums, it means that the series overall diverges.

5. How can convergence of partial sums be used to approximate the value of a series?

The convergence of partial sums can be used to approximate the value of a series by calculating the sum of a finite number of terms in the series. By adding more terms, the approximation becomes more accurate. This is particularly useful in situations where it is not possible to find an exact solution for the series.

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