If $a_n>0$ for all n, and $\sum_{n=1}^\infty a_n$ diverges, then can I prove $\sum_ {n=1}^\infty \frac{a_n}{a_n+1}$ diverges by showing that $\frac{a_n}{a_n+1}$ = $1-\frac{1}{a_n+1}$ and then use the comparison test?
Comparison test, proving divergence
0
$\begingroup$
sequences-and-series
-
0In both cases, $a_n$ does not bound $\frac{a_n}{a_n+1}$ from above, right? So how can I use the comparison test in this case? – 2012-11-14
1 Answers
4
First prove that $\frac{a_n}{a_n+1}\xrightarrow[n\to\infty]{} 0\iff a_n\xrightarrow[n\to\infty]{}0$ and then use limit comparison test to show that
$\sum_{n=1}^\infty a_n$ converges iff $\sum_ {n=1}^\infty \frac{a_n}{a_n+1}$ converges:$\displaystyle{\lim_{n\to \infty}\dfrac{\frac{a_n}{1}}{\frac{a_n}{a_n+1}}}=\lim_{n\to \infty}a_n+1=1 \neq 0.$
-
0@Alti: You are welcome. – 2012-11-14