Show that if Σ((an)/(1+an)) converges,then Σan also converges

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In summary, the sequence $\sum_{n=1}^{\infty} a_{n}$ converges if and only if the sequence $\sum_{n=1}^{\infty} \frac{a_{n}}{1+a_{n}}$ converges.
  • #1
evinda
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Hello! :)
I am given this exercise:
Let $a_{n}$ be a sequence of positive numbers.Show that the sequence $\sum_{n=1}^{\infty} a_{n}$ converges if and only if the sequence $\sum_{n=1}^{\infty} \frac{a_{n}}{1+a_{n}}$ converges.
That's what I have tried so far:
->We know that $\frac{a_{n}}{a_{n}+1}\leq a_{n}$ ,so from the Comparison Test,if the sequence $\sum_{n=1}^{\infty} a_{n}$ converges,then the sequence $\sum_{n=1}^{\infty} \frac{a_{n}}{1+a_{n}}$ also converges.
->If the sequence $\sum_{n=1}^{\infty} \frac{a_{n}}{1+a_{n}}$ converges, $\frac{a_{n}}{a_{n}+1} \to 0$,so there is a $n_{0}$ such that $\frac{a_{n}}{1+a_{n}}<\frac{1}{2} \forall n \geq n_{0}$ ...But how can I continue? :confused:
 
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  • #2
I think you have some typos then, so you're watching $a_n+1$ instead of $a_{n+1}.$
Since $a_n>0$ (this fact is very important in order these things work), we have $\dfrac{{{a}_{n}}}{{{a}_{n}}+1}\le {{a}_{n}},$ so the converse implication is trivial.

As for the other, again, since $a_n>0,$ you can use the limit comparison test for $a_n$ and $\dfrac{a_n}{1+a_n}$ to conclude.
 
  • #3
Krizalid said:
I think you have some typos then, so you're watching $a_n+1$ instead of $a_{n+1}.$
Since $a_n>0$ (this fact is very important in order these things work), we have $\dfrac{{{a}_{n}}}{{{a}_{n}}+1}\le {{a}_{n}},$ so the converse implication is trivial.

As for the other, again, since $a_n>0,$ you can use the limit comparison test for $a_n$ and $\dfrac{a_n}{1+a_n}$ to conclude.

How can I use the comparison test,to show that if the sequence $\sum_{n=1}^{\infty} \frac{a_{n}}{1+a_{n}}$ converges,$\sum_{n=1}^{\infty} a_{n}$ also converges?
 
  • #4
Since $a_n/(1+a_n) \to 0$ there is an integer $N$ so that,
$$ \tfrac{a_n}{1 + a_n} < \tfrac{1}{2} \implies a_n < 1 \text{ for all }n\geq N$$
Thus,
$$ \tfrac{a_n}{1+1} < \tfrac{a_n}{1+a_n} \text{ for all}n\geq N$$
 
  • #5
ThePerfectHacker said:
Since $a_n/(1+a_n) \to 0$ there is an integer $N$ so that,
$$ \tfrac{a_n}{1 + a_n} < \tfrac{1}{2} \implies a_n < 1 \text{ for all }n\geq N$$
Thus,
$$ \tfrac{a_n}{1+1} < \tfrac{a_n}{1+a_n} \text{ for all}n\geq N$$

Great..I undertand!Thank you! :)
 

FAQ: Show that if Σ((an)/(1+an)) converges,then Σan also converges

1. What does Σ((an)/(1+an)) represent in this scenario?

The expression Σ((an)/(1+an)) represents a series, where each term is the ratio of an to (1 + an).

2. How can it be proven that if Σ((an)/(1+an)) converges, then Σan also converges?

This can be proven using the Comparison Test for series, which states that if 0 ≤ an ≤ bn for all n, and Σbn converges, then Σan also converges.

3. Is the converse of this statement also true?

No, the converse of this statement is not true. Just because Σan converges does not necessarily mean that Σ((an)/(1+an)) will converge.

4. Can this result be applied to all series with positive terms?

Yes, this result can be applied to all series with positive terms. As long as the series satisfies the conditions of the Comparison Test, the convergence of Σ((an)/(1+an)) implies the convergence of Σan.

5. Are there any other methods for proving this result?

Yes, there are other methods for proving this result, such as the Ratio Test or the Root Test. However, the Comparison Test is often the most straightforward and efficient method to use in this scenario.

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