Convergence of the difference between the Nth harmonic number and Ln(N) My question is how would you go about proving that the sequence given by the difference of the nth harmonic number and the natural log of that number converges. 
 A: Observe that
$$(\forall k \in \mathbb{N}) \quad \frac{1}{k} - \int_k^{k+1} \frac{1}{x} dx > 0$$
Show that:
$$(\forall k \in \mathbb{N}) (\exists c_k \in (k,k+1)) \quad \frac{1}{k} - \int_k^{k+1} \frac{1}{x} dx = \frac{1}{2 c_k^2}$$
(This is the error estimate for the rectangle rule, which is proven with the mean value theorem.)
Conclude that
$$(\forall k \in \mathbb{N}) \quad 0 < \frac{1}{k} - \int_k^{k+1} \frac{1}{x} dx < \frac{1}{2k^2}$$
Recall:
$$H_n - \ln(n+1) = \sum_{k=1}^n \frac{1}{k} - \int_1^{n+1} \frac{1}{x} dx$$ 
So by the above, $H_n - \ln(n+1)$ is an increasing sequence which is bounded above by $\sum_{k=1}^\infty \frac{1}{2k^2} < \infty$, so it converges. 
Finally notice:
$$H_n - \ln(n) = H_n - \ln(n+1) + \ln(n+1) - \ln(n) \\\
\ln(n+1)-\ln(n)=\ln(1+1/n) \leq 1/n$$ so $H_n - \ln(n)$ also converges.
A: Let $H_n$ denote the partial sum of the harmonic series.
Show that $H_n - \log(n)$ is a $decreasing$ sequence that is bounded below.
(Note that $H_1 - \log(1)  = 1$, $H_2 - \log(2) \approx 0.8069$, and the sequence is not increasing.)
We have
$$\frac1{k+1} < \int_{k}^{k+1}\frac{dx}{x}=\log(k+1)-\log(k) <\frac1{k.}$$
Then for $n > 1$
$$\sum_{k=1}^{n-1}\frac1{k+1}<\log n <\sum_{k=1}^{n-1}\frac1{k},\\H_n-1<\log n< H_n-\frac1{n}.$$
Hence,
$$0 < \frac1{n} < H_n -\log n \leq 1.$$
So $H_n - \log n $ is bounded between $0$ and $1$ for all $n \in \mathbb{N}$.
Also
$$\frac1{n+1} < \log(n+1)-\log(n),\\\frac1{n+1} - \log(n+1)< -\log(n),\\ \sum_{k=1}^{n}\frac1{k} + \frac1{n+1} - \log(n+1)<\sum_{k=1}^{n}\frac1{k}  - \log(n).$$
Hence $H_{n+1} - \log(n+1) < H_n - \log(n)$ and the sequence is decreasing and convergent to a limit between $0$ and $1$.
