Limit of a recursive sequence with an independent term that goes to 0 I have a recursive sequence where the first element is $a_{1} = 1$ and then $a_{n+1}= \frac12a_{n} + \frac{1}{n+1}$.
The first three terms are $a_{1} = 1 > a_{2} = \frac56 > a_{3} = \frac23$ so it seems reasonable to try to prove that the sequence is decreasing and then use the monotonicity of the sequence to try to prove that it has a finite or infinite limit. (This is my usual approach to recursive sequences.)
However, I don't know how to deal with the term $\frac1{n+1}$. Clearly $\lim \frac1{n+1} = 0$, but I don't know if I can erase it and solve $a_{n+1} = \frac12 a_{n} < \frac12 a_{n-1} = a_{n}$ to prove the monotonicity.

First prove a Lemma: $a_n > \frac{2}{n+1}$ for $n \geq2 $
When $n=2$, $a_2 = 5/6 > 2/3$
Now suppose $a_n > \frac{2}{n+1}$
then $a_{n+1} = \frac{a_n}{2}+ \frac{1}{n+1} > (\frac{1}{n+1})+\frac{1}{n+1} = \frac{2}{n+1} $
By induction, $a_n > \frac{2}{n+1}$ for all n 
Now we can prove that $a_n$ is decresing.
$a_{n+1}-a_n = \frac{a_n}{2}+ \frac{1}{n+1}-a_n = \frac{-a_n}{2}+ \frac{1}{n+1} < (-\frac{1}{n+1}) + \frac{1}{n+1} = 0$
so $a_n$ is decresing. By our Lemma, we also have $a_n \geq 0$, so we have a decresing sequence bounded below, thus it converges.
To find the limit, let $lim_{n\to \infty}  a_n = L$. By the recursive relation,
$L = L/2$ as $n\to \infty$
so $L = 0$
$lim_{n\to \infty} a_n = 0$
