How to find the limit for the quotient of the least number $K_n$ such that the partial sum of the harmonic series $\geq n$ Let $$S_n=1+1/2+\cdots+1/n.$$ Denote by $K_n$ the least subscript $k$ such that $S_k\geq n$. Find the limit $$\lim_{n\to\infty}\frac{K_{n+1}}{K_n}\quad ?$$
 A: We know that $H_n=\ln n + \gamma +\epsilon(n)$, where $\epsilon(n)\approx \frac{1}{2n}$ and in any case $\epsilon(n)\rightarrow 0$ as $n\rightarrow \infty$.  If $m=H_n$ we may as a first approximation solve as $n=e^{m-\gamma}$.  Hence the desired limit is $$\lim_{m\rightarrow \infty} \frac{e^{m+1-\gamma}}{e^{m-\gamma}}=e$$
For a second approximation, $m=\gamma + \ln n +\frac{1}{2n}=\gamma+\ln n+\ln e^{\frac{1}{2n}}=\gamma+\ln ne^{\frac{1}{2n}}$.  This may be rearranged as $ne^{\frac{1}{2n}}=e^{m-\gamma}$.  This has solution $$n=-\frac{1}{2W(-e^{\gamma-m}/2)}$$
where $W$ is the Lambert function.  Hence the desired limit is now $$\lim_{m\rightarrow \infty}\frac{W(-e^{\gamma-m}/2)}{W(-e^{\gamma-m-1}/2)}=e$$
Although not a proof, this is compelling enough that I'm not going to think about the next error term.
A: Consider that, for $p>k>1$,
$$
S_k+\int_{k+1}^{p}\frac1xdx\leq S_{p} \leq S_k+\int_{k}^{p-1}\frac1xdx
$$
Now, consider $S_{K_n}=n+\epsilon_n$, where $0<\epsilon_n<\frac1{K_n}$ for $n>1$. If we let $k=K_n$ and $p=K_{n+1}$, then we have
$$
S_{K_n}+\log(K_{n+1})-\log(K_n+1)\leq S_{K_{n+1}}\leq S_{K_n}+\log(K_{n+1}-1)-\log(K_n)
$$
or
$$
n+\epsilon_n+\log\left(\frac{K_{n+1}}{K_n+1}\right)\leq n+1+\epsilon_{n+1}\leq n+\epsilon_n+\log\left(\frac{K_{n+1}-1}{K_n}\right)
$$
As such, we can write that
$$
\log\left(\frac{K_{n+1}}{K_n+1}\right)\leq 1+\epsilon_{n+1}-\epsilon_n\leq \log\left(\frac{K_{n+1}-1}{K_n}\right)
$$
or
$$
\frac{K_{n+1}}{K_n+1}\leq e^{1+\epsilon_{n+1}-\epsilon_n}\leq \frac{K_{n+1}-1}{K_n}
$$
Now, we may write the left inequality as
$$
K_{n+1} \leq e^{1+\epsilon_{n+1}-\epsilon_n}(K_n+1)
$$
and the right inequality as
$$
e^{1+\epsilon_{n+1}-\epsilon_n}K_n+1\leq K_{n+1}
$$
and so
$$
e^{1+\epsilon_{n+1}-\epsilon_n}+\frac1{K_n}\leq \frac{K_{n+1}}{K_n} \leq e^{1+\epsilon_{n+1}-\epsilon_n}(1+\frac1{K_n})
$$
Taking the limit as $n\to\infty$, we note that $\lim_{n\to\infty} \epsilon_n=0$ and $\lim_{n\to\infty} \frac1{K_n}=0$, and so
$$
e\leq \lim_{n\to\infty} \frac{K_{n+1}}{K_n} \leq e
$$
Therefore,
$$
\lim_{n\to\infty} \frac{K_{n+1}}{K_n} = e
$$
