Question: I would like some help with the following sequence:
$b\_{n}=b\_{n-1}-\frac{1}{2n}+\frac{1}{2n+1}$ and $b\_1=\frac{5}{6}$, then $b\_2=\frac{5}{6}-\frac{1}{20}$. (this second part is edited, I left out $\frac{5}{6}$, sorry).

I've already proved the sequence is monotonically decreasing and therefore the first term will be an upperbound. In this case it is clear that $\frac{5}{6}$ is an upperbound. Now I've to prove that it has a lowerbound. I feel that $M=0$ would suffice but I get stuck right here:
Let $M \in \mathbb{R}$ be an upperbound for $b\_{n}$. Then $|b\_{n}-M|=|b\_{n}-0|=|b\_n|<\epsilon$, with $\epsilon>0$. 

I am stuck at this point. What N to pick to make this work?

Thanks

Answer: 
Observe that $b\_n-b\_{n-1}=\dfrac{1}{2n+1}-\dfrac{1}{2n}$ gives us $$b\_{n}-b\_1=-\dfrac14+\dfrac15-\dfrac16+\dfrac17-\dfrac18+\cdots -\dfrac{1}{2n}+\dfrac{1}{2n+1}$$ and by alternating series test RHS is convergent.
Note that
$$b\_n=b\_{n-1}-\frac{1}{2n}+\frac{1}{2n+1}=b\_{n-1}-\frac{1}{2n(2n+1)}.$$
Therefore,
$$b\_n=b\_1-\sum\_{k=2}^n \frac{1}{2k(2k+1)}.$$
Since the series
$$\sum\_{k=2}^{\infty} \frac{1}{2k(2k+1)}=\sum\_{k=2}^{\infty} \frac{1}{4k^2+2k}$$
consists of positive terms and is convergent (say, by the Limit Comparison Test with $\sum\frac{1}{k^2}$), it has a finite sum
$$\sum\_{k=2}^{\infty} \frac{1}{2k(2k+1)}=S, \quad \text{and for any $n$:} \quad \sum\_{k=2}^n \frac{1}{2k(2k+1)}<S.$$
Therefore, 
$$b\_n=b\_1-\sum\_{k=2}^n \frac{1}{2k(2k+1)}>b\_1-S.$$