Prove that a sequence is bounded 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
 A: 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.
A: 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, 
$$b_n=b_1-\sum_{k=2}^n \frac{1}{2k(2k+1)}>b_1-S.$$
A: 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.
