Does $\lim_{k \to \infty}[(\sum_{n=1}^{k}\frac{1}{n})-\ln k]$ exists? Let $k>2$, is a natural number, does $\lim_{k \to \infty}[(\sum_{n=1}^{k}\frac{1}{n})-\ln k]$ exists? If $k \to \infty$, then I think no, because $\ln k$ will go to infinite. And $\sum \frac{1}{n}$ diverge too, and so it's $\infty - \infty$. But someone say the limit exists? and it's positive. How?

Note that
$$
 - \log k =  - \log \left( {\frac{2}{1} \cdots \frac{{k - 1}}{{k - 2}}\frac{k}{{k - 1}}} \right) =  - \sum\limits_{n = 2}^k {\log \left( {\frac{n}{{n - 1}}} \right)}  = \sum\limits_{n = 2}^k {\log \left( {1 - \frac{1}{n}} \right)} .
$$
Thus,
$$
\sum\limits_{n = 1}^k {\frac{1}{n}}  - \log k = 1+\sum\limits_{n = 2}^k {\left( {\frac{1}{n} + \log \left( {1 - \frac{1}{n}} \right)} \right)} .
$$
Now if $0<x\leq \frac{1}{2}$, then
$$
\left| {x + \log (1 - x)} \right| = \left| { - \int_0^x {\frac{t}{{1 - t}}dt} } \right| \le 2\int_0^x {tdt}  = x^2 .
$$
Therefore,
$$
\left| {\sum\limits_{n = 2}^k {\left( {\frac{1}{n} - \log \left( {1 - \frac{1}{n}} \right)} \right)} } \right| \le \sum\limits_{n = 2}^k {\left| {\frac{1}{n} - \log \left( {1 - \frac{1}{n}} \right)} \right|}  \le \sum\limits_{n = 2}^k {\frac{1}{{n^2 }}}  < \sum\limits_{n = 2}^\infty  {\frac{1}{{n^2 }}}  < \infty .
$$
Consequently, the sequence
$$
1 + \sum\limits_{n = 2}^k {\left( {\frac{1}{n} - \log \left( {1 - \frac{1}{n}} \right)} \right)}  = \sum\limits_{n = 1}^k {\frac{1}{n}}  - \log k
$$
is convergent.
