$a_{n}=1+\frac{1}{2}+\cdots+\frac{1}{n}-\log(n)$ converges 
Prove that $$a_{n}=1+\frac{1}{2}+\cdots+\frac{1}{n}-\log(n)\  \text{ converges.}$$

any hints. 
I am trying to upper and lower bound it by a sequence that converges to the same number.
 A: Hint:
$$
\log k - \log(k-1) = \int_{k-1}^k \frac{dt}t$$
detail:
$$
\log k - \log(k-1) = \int_{k-1}^k \frac{dt}t
\\
\int_k^{k+1}\frac{dt}t <\frac 1k < \int_{k-1}^k \frac{dt}t 
\\
\log(n+1) - \log 2 < \sum_{k=2}^n \frac 1k <\log n
\\
1-\log 2<1-\log 2 + \log(n+1)-\log n < \sum_{k=1}^n \frac 1{k} -\log n < 1
$$so the sequence is bounded.
$$
a_{n}-a_{n-1} = \frac 1n  - \log{n}+ \log(n-1)
=\frac 1n - \int_{n-1}^n\frac{dt}t<0
$$so the sequence is monotonic. Hence it is convergent.
another solution:
Let $$b_n = 1+\frac 12+\cdots+\frac 1n -\log(n+1)$$
Then $$b_n - b_{n-1} = \frac 1n  - \log(n+1)+ \log n >0$$ so $(b_n)$ is increasing.
$$
a_n - b_n = \log(n+1) - \log n \to 0
$$
so, as $(a_n)$ is decreasing,
$(a_n)$ and $(b_n)$ are both convergent.
A: $$
\begin{align}
& \Big( \log 1 - \log 2 \Big) + \Big( \log 2 - \log 3 \Big) + \Big( \log 3 - \log 4 \Big) + \cdots\\
& \cdots+\Big( \log(n-3) - \log (n-2) \Big) +\Big( \log(n-2) -\log(n-1) \Big)+\Big(\log(n-1)-\log n\Big)
\end{align}
$$
Everything above cancels except the first term and the last, and the first term, $\log 1$, is $0$, so the whole thing adds up to $-\log n$.
If we now add $1/n$ to the last term, and $1/(n-1)$ to the term before that, and $1/(n-2)$ to the term before that, and so on, we will add $1/2$ to the first term, and then just add $1$ and we've got just your sum.
The typical term is $\log k - \log (k+1) + \dfrac 1 {k+1}$.  This is the same as
$$
\frac1{k+1} - \left( \log(k+1) - \log k\right) = - \left(\int_k^{k+1} \frac{dx}{x} - \frac 1 {k+1} \right).
$$
Next:
$$
\begin{align}
& \int_k^{k+1} \frac{dx}{x} - \frac 1 {k+1} = \int_0^1 \frac{dx}{x+k} - \frac{1}{1+k} \\[8pt]
= {} & \text{area of that part of the square $0<x<1,\  0<y<1$ that is} \\
& \text{above the line $y=1/(k+1)$ and below the curve $y=1/(x+k)$.}
\end{align}
$$
Now notice that those parts of the square do not overlap!!  That means they're total area must be no bigger than the area of the whole square, i.e. their sum converges.
