Convergence of $\sum_{i=0}^n \frac{(\log(n+i)-\log n)^2}{n+i}$ Define a sequence $S_n$ of real numbers by $\sum_{i=0}^n \frac{(\log(n+i)-\log n)^2}{n+i}$ .
Does the $\lim_{n\to \infty}S_n$ exist? If so, compute the value of this limit.
I am getting two different answers if I use Cauchy theorem and if i convert this to integral form. Need help.
Attempt:
Let $f_n$ = $n\frac{(\log(n+n)-\log n)^2}{n+n}$ = $\frac{(log2)^2}2$ as n tends to $\infty$
Therefore by Cauchy Theorem $\sum \frac{f_1 + f_2+...+f_n}n$ = $S_n$  and $\lim_{n\to \infty}S_n$ = $\lim_{n\to \infty}f_n$ =  $\frac{(log2)^2}2$
 A: Well, $\log(i+n)-\log n = \log\left(1+\frac{i}{n}\right)$ is non-negative but less than $\frac{i}{n}$, and:
$$ \sum_{i=0}^{n}\frac{i^2}{n^2(n+i)}=\sum_{i=0}^{n}\left(\frac{1}{n+i}-\frac{n-i}{n^2}\right)=H_{2n}-H_{n-1}-\frac{n+1}{n}+\frac{n(n+1)}{2n^2}$$
is convergent to $\log(2)-\frac{1}{2}$. Then, by a Riemann sum argument:
$$ \lim_{n\to +\infty}\frac{1}{n}\sum_{i=0}^{n}\frac{\log^2\left(1+\frac{i}{n}\right)}{1+\frac{i}{n}}=\int_{0}^{1}\frac{\log^2(1+x)}{1+x}\,dx=\left.\frac{1}{3}\log^3(1+x)\right|_{0}^{1}=\color{red}{\frac{\log^3 2}{3}}.$$
A: I can prove $S_n$ is bounded, but for convergence you also need monotonicity (which I believe you have, since going from $n$ to $n+1$ you remove large terms and add smaller terms, so it should be decreasing). Anyways, observe that
$$
S_n \leq \frac{\ln^2 2}{n} + \sum_{i=1}^n \frac{(\ln(2n-1)-\ln n)^2}{n+i}\\
= \frac{\ln^2 2}{n} + \sum_{i=1}^n \frac{\left(\ln\left(1+1-\frac{1}{n}\right)\right)^2}{n+i}\\
\leq \frac{\ln^2 2}{n} + \sum_{i=1}^n \frac{\left(1-\frac{1}{n}\right)^2}{n+i}\\
= \frac{\ln^2 2}{n} + \frac{(n-1)^2}{n^2}\sum_{i=1}^n \frac{1}{(n+i)}
$$
The first part approaches zero, so if we prove the second part is bounded, we're done. Also, notice that the coefficient in front of the sum approaches 1, so the only issue is whether or not this quantity converges
$$
c_n=\sum_{i=1}^{n}\frac{1}{n+i} = \sum_{i=1}^{2n} \frac{1}{i} - \sum_{i=1}^n \frac{1}{i}
$$
But adding and subtracting $\ln(2n)$ we get
$$
c_n = \sum_{i=0}^{2n} \frac{1}{i} - \ln(2n) - \sum_{i=0}^n \frac{1}{i} + \ln(n) + \ln 2 \to \gamma - \gamma + \ln 2 = \ln 2
$$
where $\gamma$ is the Euler-Mascheroni constant. Therefore, the upper bound converges, and therefore is bounded. Hence, $S_n$ is bounded.
