Question Sequences and Series Prove that the sequence $(a_n)_{n \geq 1}$ defined by $$ a_n= \sum_{k=1}^{n} \frac{1}{k} - \ln(n+1)$$
 is increasing and bounded. It is on the study guide for my final exam, which is tomorrow so I am trying to figure it out. Thanks
Can someone please help?
 A: Draw the curve y=1/x and look at how $\sum 1/k$ and $ln(n+1)$ are represented geometrically.  $ln(n+1)$ is the area under the curve from 1 to n+1, and $\sum 1/k$ is the total of a bunch of rectangles of width 1, each with a triangular(ish) extra bit above the curve.  Each time you bump n up by 1, you add one more little triangle to the difference, so it's increasing.
Then to show the total difference is bounded, imagine taking all those triangles and shifting them all the way to the left so they're against the y-axis.  They'll all be contained in a 1x1 square without overlapping.
A: We have, for any  $n \geq  1$, $$a_{n+1} - a_{n}=  \sum_{k=1}^{n+1}  \frac{1}{k} - \ln(n+2) -\sum_{k=1}^{n}  \frac{1}{k} + \ln(n+1) =  \frac{1}{n+1} - \ln(\frac{n+2}{n+1})$$ $$ = \frac{1}{n+1} - \ln(\frac{1}{n+1}+1) $$
Now consider the function $ f(x) = x-\ln(1+x)$ defined on $[0,1]$. we have $ f'(x) = 1- \frac{1}{1+x}  \geq 0 ,\; \forall x \in [0,1]$ , thus $f$ is increasing on $[0,1]$, hence  $f(x) \geq f(0) \; \forall x \in [0,1] $, so that  $ x-\ln(1+x) \geq 0 \; \forall x \in [0,1] $, in particular for $x=\frac{1}{n+1} , n \geq 1 $, we get   $ \frac{1}{n+1} - \ln(\frac{1}{n+1}+1) \geq 0 $. Thus $a_{n+1} - a_{n} \geq 0$ , therefore $a_n $ is increasing . 
A: The part that $a_n$ is increasing is already done. We focus on proving $a_n$ is bounded above. We have:
$a_n = 1 + \dfrac{1}{2} + \dfrac{1}{3} + \cdots + \dfrac{1}{n} - \ln(n+1) = 1 + \dfrac{1}{2} + \cdots + \dfrac{1}{n} - \ln n - \ln\left(1+\frac{1}{n}\right) = 1 + \dfrac{1}{2} + \cdots + \dfrac{1}{n-1} - \ln n+ \left(\dfrac{1}{n} - \ln\left(1+\frac{1}{n}\right)\right)$. Apply the MVT for the function $f(x) = x - \ln(1+x)$ on $(0,x)$ we have: $x - \ln(1+x) = f'(c)\left(x-0\right) = \dfrac{cx}{1+c} < cx, 0 < c < x$. Let $x = \dfrac{1}{n} \Rightarrow \dfrac{1}{n} - \ln(1+\frac{1}{n}) < c\dfrac{1}{n} < \dfrac{1}{n^2}$. By repeating this pattern back to $\ln 2$ we have that $a_n < 1 + \dfrac{1}{2^2} + \cdots + \dfrac{1}{n^2} < \dfrac{\pi^2}{6}$, proving boundedness of $a_n$.
