Check that $a_n = \sum_{k=1}^n \frac{1}{n+k}$ is bounded from above by $\frac 34$ Any ideas on how to check that
$
a_n = \sum_{k=1}^n\frac{1}{n+k} \le \frac 34\,?
$
It's pretty easy to see that the sequence is monotonic and has an almost trivial upper bound of $1$ (and a trivial lower bound of $1/2$).
I solved it several years ago but can't recall my solution. I do remember I scratched my head for a while, though. I'm quite sure one needs to get an upper bound through some algebraic manipulation to get $\sum_{k=2}^\infty \frac {1}{k^2}$ (well, some polynomial of order 2 in $k$) but can't recover the proper way to do it (without relying on integral calculus).
Wolframalpha suggests that $\lim_n a_n = \ln(2)$, which can be easily seen by integration. Is there a `more' elementary way to see it? My guess would be no.

If you already proved that $H_{2n}-H_n$ is increasing with respect to $n$ you are essentially done.
Here $H_n$ is the $n$-th harmonic number, $\sum_{k=1}^{n}\frac{1}{k}$, and by Riemann sums
$$\lim_{n\to +\infty}\left(H_{2n}-H_n\right)=\lim_{n\to +\infty} \sum_{k=1}^{n}\frac{1}{k+n}=\lim_{n\to +\infty}\frac{1}{n}\sum_{k=1}^{n}\frac{1}{1+\frac{k}{n}}=\int_{0}^{1}\frac{dx}{1+x}=\log 2.$$
As an alternative,
$$ H_{2n}-H_n = \sum_{k=1}^{2n}\frac{(-1)^{k+1}}{k}$$
is clearly related with the Taylor series of $\log(1+x)$ at the origin.
Thus it is enough to prove $\log(2)<\frac{3}{4}$. We can do much better. By computing a polynomial remainder we have
$$ \int_{0}^{1}\frac{x^4(1-x)^4}{1+x}\,dx =-\frac{621}{56}+16\log 2,$$
where the integrand function is positive but bounded by $\frac{1}{4^4}$ on $[0,1]$. It follows that:
$$ H_{2n}-H_n \leq \log(2) < \color{red}{\frac{39}{56}}.$$
