So I'm working on trying to show that $$\sum_{k=2}^{n-1}\frac{1}{\log(k)}-\int_{2}^n\frac{1}{\log(x)}$$ converges to a real number less than $\log(2).$ It's not hard to show the sequence is monotone for $$\sum_{k=2}^{n-1}\frac{1}{\log(k)}-\int_{2}^n\frac{1}{\log(x)}=\sum_{k=2}^n\int_{k}^{k+1}\left[\frac{1}{\log(k)}-\frac{1}{\log(x)}\right],$$ which is increasing w.r.t. $n$ as $1/\log(k)>1/\log(x)$ for $x>k.$ Now we know it's supposed to converge to a real number less than $\log(2),$ so showing $\log(2)$ is an upper bound will gives us our desired result by order limit properties and Monotone Convergence. Now I'm not totally sure what $\int_{k}^{k+1}1/\log(x)$ is, so I've been trying to work out a suitable upper bound. So far I've concluded that $$\sum_{k=2}^{n-1}\frac{1}{\log(k)}-\int_{2}^n\frac{1}{\log(x)}<\int_{2}^3\left(\frac{1}{\log(2)}-\frac{1}{\log(x)}\right)+\sum_{k=3}^n\frac{(k+1)\ln(1+1/k)-1}{\ln(k)^2}.$$ Since $\log(2)=1/(1\cdot2)+1/(3\cdot 4)+...$ I've been trying to show that $$\sum_{k=2}^{n-1}\frac{1}{\log(k)}-\int_{2}^n\frac{1}{\log(x)}\leq\sum_{k=2}^{n-1}\frac{1}{(2(k-1)-1)\cdot(2(k-1))}.$$ Geogebra suggests this approach might work as according to the graphs we should have $$\sum_{k=3}^n\frac{(k+1)\ln(1+1/k)-1}{\ln(k)^2}<\sum_{k=3}^{n-1}\frac{1}{(2(k-1)-1)\cdot(2(k-1))}.$$ If we can prove the above inequality, then the problem reduces to a single term, that is, it reduces to showing $$\int_{2}^3\left(\frac{1}{\log(2)}-\frac{1}{\log(x)}\right)<\frac{1}{2}.$$
Problem is, I don't see how to prove either of the last two inequalities. Also, maybe there is a much simpler way to go about this, that I just don't see. Any help would be lovely. Thank you.
There is indeed a simpler way to finish off the question. Note that $$ \frac1{\log k} - \int_k^{k+1} \frac{dx}{\log x} < \frac1{\log k} - \int_k^{k+1} \frac{dx}{\log(k+1)} = \frac1{\log k} - \frac1{\log(k+1)}. $$ Summing both sides over $k$ yields \begin{align*} \sum_{k=2}^{n-1} \frac1{\log k} - \int_2^n \frac{dx}{\log x} &= \sum_{k=2}^{n-1} \bigg( \frac1{\log k} - \int_k^{k+1} \frac{dx}{\log x} \bigg) \\ &< \sum_{k=2}^{n-1} \bigg( \frac1{\log k} - \frac1{\log(k+1)} \bigg) = \frac1{\log 2}-\frac1{\log n}. \end{align*}
Geometrically this argument is just the observations that the curvy triangles that represent the difference between the (left Riemann) sum and the area under the graph can be stacked vertically to fit inside a $1\times\big(\frac1{\log2}-\frac1{\log n}\big)$ rectangle.