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

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,723
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:
 
Last edited:

Krizalid

Active member
Feb 9, 2012
118
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.
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,723
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???
 

ThePerfectHacker

Well-known member
Jan 26, 2012
236
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,
$$ \tfrac{a_n}{1+1} < \tfrac{a_n}{1+a_n} \text{ for all}n\geq N$$
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,723
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!!! :)