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.
Rewrite the relation to the form below:

$$b\_n-b\_{n-1}=\dfrac 1{2n}+\dfrac 1{2n+1}-\dfrac 1n$$

By summing this telescoping stuff we get 

$b\_n-b\_0=\sum\limits\_{k=2}^{2n+1}\dfrac 1k-\sum\limits\_{k=1}^{n}\dfrac 1{k}$

Since $b\_0=1$ (deduced from $b\_1=\frac 56$) we can reintroduce it into the first sum and we get

$$b\_n=\sum\limits\_{k=n+1}^{2n+1}\dfrac 1k$$

Now we have a limit by noticing it is a Riemann sum

$\displaystyle b\_n=\sum\limits\_{k=0}^{n}\dfrac 1{n+1+k}=\dfrac 1{(n+1)}\sum\limits\_{k=0}^{n}\dfrac 1{1+\left(\frac k{n+1}\right)}\to \int\_0^1 \dfrac{\mathop{dx}}{1+x}=\ln(2)$

This is a stronger result than just the lower bound asked, but notice that it suffice to say it is a Riemann sum and the integral is obviously finite since $\frac 1{1+x}$ is continuous on $[0,1]$. So we get the convergence anyway, even in the eventuallity we were unable to calculate this integral. Finally since every convergent sequence is bounded we have our result.

However I feel that it is not much more difficult to go for this limit directly, than invoking the alternating series test theorem or the comparison test with $\sum \frac 1{k^2}$ in the other answers.