Proving Series Divergence: Convergence of a_n w/o Explicit Formula

In summary, to show that the series diverges, we can use the comparison test and bound the series from below by showing that for a_n > 1, the series \frac{a_n}{1+a_n} is greater than \frac{1}{2}, and for a_n \leq 1, the series \frac{a_n}{1+a_n} is greater than or equal to \frac{a_n}{2}. This shows that the series \frac{a_n}{1+a_n} is always greater than some positive constant, and therefore diverges since \sum a_n diverges.
  • #1
CrusaderSean
44
0
Given [tex]a_{n} > 0[/tex] and [tex]\sum a_{n}[/tex] diverges, show that [tex]\sum \frac{a_{n}}{1+a_{n}}[/tex] diverges.
Since I don't have an explicit form for the series, I can't apply any of the standard tests. I'm not sure where to start on this problem. I know the criteria for convergence/divergence, namely tail end of series has to converge or cauchy criterion condition. But I don't see how that helps without knowing what series looks like. Please steer me in the right direction.
 
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  • #2
If a series diverges, what happens to to its reciprocal ?
 
  • #3
I would say reciprocal converges, but apparently it's not enough to bring original series to convergence... I thought about this a little more and I think I'll analyze it based on how [tex]a_{n}[/tex] diverges. that is, does it go to zero, constant, or infinity as n goes to infinity and try to bound the reciprocal from below to show series diverges.
 
  • #4
For [itex]a_n>1[/itex]:

[tex]\frac{a_n}{1+a_n}>\frac{1}{2}[/tex]

For [itex]a_n\leq 1[/itex]:

[tex]\frac{a_n}{1+a_n}\geq\frac{a_n}{2}[/tex]
 

1. What is the definition of a divergent series?

A divergent series is a mathematical series in which the terms do not approach a finite limit as the number of terms increases. In other words, the sum of the terms in a divergent series does not have a finite value.

2. How do you prove that a series is divergent?

There are several methods for proving the divergence of a series. One way is to show that the terms of the series do not approach zero as the number of terms increases. Another method is to use a comparison test, where the series is compared to a known divergent series. Additionally, the integral test, ratio test, and root test can also be used to prove divergence.

3. What is the importance of proving series divergence?

Proving series divergence is important because it helps us determine the behavior of a series and whether or not it has a finite sum. Divergent series have no finite sum, which means they cannot be used in calculations or equations that require a finite value.

4. Can a series be both convergent and divergent?

No, a series cannot be both convergent and divergent. A series can either have a finite sum, making it convergent, or it can have an infinite sum, making it divergent. It cannot have both properties simultaneously.

5. How can the convergence of a series be determined without an explicit formula for the terms?

If an explicit formula for the terms of a series is not available, the convergence or divergence of the series can still be determined by using various convergence tests. These tests compare the terms of the series to known series and can determine the behavior of the series without an explicit formula.

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