Approximation of harmonic sum as a multiple of $\ln n$

Question

It is well-known that the harmonic series $\sum _{i=0}^{n-1} \frac{1}{1+i}$ could be approximated to the natural logarithm $\ln$. Now, let $c$ be an arbitrary positive real number. Could one deduce that for a big enough $n$ the sum $\sum_{i=0}^{n-1} \frac{1}{i+e}$ is approximated as (or is less than) $\frac{1}{2c}\ln n$? Here, $e$ is the Euler base of natural logarithm.\ Thanks for any help!

Answer 1

$$\sum_{i=0}^{i=n}\frac{1}{1+c\:i}=\frac{1}{c}\left(\psi(n+1+\frac{1}{c})-\psi(\frac{1}{c})\right)\qquad n>0$$

$\psi(x)$ is the digamma function. https://mathworld.wolfram.com/DigammaFunction.html

Asymptotic series :

$$\psi(z+1)\sim \ln(z)+\frac{1}{2z}-\sum_{k=1}^\infty \frac{B_{2k}}{2k\:z^{2k}}$$ $B_{2k}$ are the Bernoulli numbers : https://mathworld.wolfram.com/BernoulliNumber.html

Approximate at large $n$: $$\psi(z+1)\sim \ln(z)+\frac{1}{2z}+...\quad\implies\quad \psi(n+1+\frac{1}{c})\sim \ln(n+\frac{1}{c})+\frac{1}{2(n+\frac{1}{c})}+...$$ $$\psi(n+1+\frac{1}{c})\sim \ln(n)+(\frac{2-c}{2c})\frac{1}{n}+...$$

$$\sum_{i=0}^{i=n\to\infty}\frac{1}{1+c\:i}\sim \frac{1}{c}\left(\ln(n)-\psi(\frac{1}{c})+(\frac{2-c}{2c})\frac{1}{n}\right)+...$$

$$\sum_{i=0}^{i=n\to\infty}\frac{1}{\alpha+i}\sim \ln(n)-\psi(\alpha)+(\alpha-\frac{1}{2})\frac{1}{n}+...$$

Answer 2

Hint.

Let $S(n)=-\ln n+\sum_{i=0}^{n-1}1/(e+i).$ Then $$S(n+1)-S(n)=1/(e+n)-\ln (1+1/n)=B(n)-C(n)$$ where $B(n)=1/n-\ln (1+1/n)=1/2n^2-1/3n^3+...$

and $C(n)=1/n-1/(e+n)=e/(n^2+ne).$

Observe that $$S(n+1)=S(1)+\sum_{j=1}^{n+1} S(j+1)-S(j)=S(1)+\sum_{j=1}^{n+1}B(j)+C(j).$$ Observe that $\lim_{n\to\infty}\sum_{j=1}^{n+1}B(j)$ and $\lim_{n\to\infty}\sum_{j=1}^{n+1}C(j)$ both exist in $\Bbb R$ because $0<B(n)<1/2n^2$ and $0<C(n)<e/n^2$.