Show that $\int_{0}^{n} \left (1-\frac{x}{n} \right ) ^n \ln(x) dx = \frac{n}{n+1} \left (\ln(n) - 1 - 1/2 -...- 1/{(n+1)} \right )$ The problem is stated as:

Show that $\int_{0}^{n} \left (1-\frac{x}{n} \right ) ^n \ln(x) dx = \frac{n}{n+1} \left (\ln(n) - 1 - 1/2  -...- 1/{(n+1)} \right )$

My attempt
First of all, we make the substitution $1-\frac{x}{n} = t$, we then have that the integral can be rewritten as:
$\int_{1}^{0} -n t^n \ln(n(1-t)) dt = \int_{0}^{1} n t^n \ln(n(1-t)) dt$
Using logarithmic laws, we can split the integral into two seperate ones as follows:
$\int_{0}^{1} n t^n \ln(n(1-t)) dt = \int_{0}^{1} n t^n \ln(n) dt + \int_{0}^{1} n t^n \ln(1-t) dt$
We calculate each integral from the sum above:
$ I_1 := \int_{0}^{1} n t^n \ln(n) dt = \frac{n}{n+1}\ln(n)$
$ I_2 := \int_{0}^{1} n t^n \ln(1-t) dt = -n\int_{0}^{1} t^n \sum_{k=1}^{\infty}\frac{t^k}{k} dt$
Since the radius of convergence of $\sum_{k=1}^{\infty}\frac{t^k}{k}$ is 1, and we are integrating from $0$ to $1$, we can interchange the order of limit operations. Meaning, we can calculate the integral first.
$ I_2 = -n\sum_{k=1}^{\infty}\int_{0}^{1}\frac{t^{(n+k)}}{k} dt = -\sum_{k=1}^{\infty} \frac{n}{k(n+k+1)} = \frac{-n}{n+1} \sum_{k=1}^{\infty} \frac{n+1}{k(n+k+1)}$
Using partial fraction decomposition, we have that $I_2$ can be written as:
$\frac{-n}{n+1}\sum_{k=1}^{\infty} \frac{n+1}{k(n+k+1)} = \frac{-n}{n+1} \sum_{k=1}^{\infty} \frac{1}{k} + \frac{n}{n+1}\sum_{k=1}^{\infty} \frac{1}{n+k+1}$
Putting it all together we get:
$I_1 + I_2 = \frac{n}{n+1} \left ( \ln(n) -  \sum_{k=1}^{\infty} \frac{1}{k} + \sum_{k=1}^{\infty} \frac{1}{n+k+1} \right )$
Which is indeed close the the result sought, however, I don't really know what to do with the last sums, and why I did wrong in choosing $\infty$ as an upper limit in the summation. I see that the sum of $1/k's$ diverge, but how can I avoid this?
Thank you for any help that could help me complete the last step of this problem.
 A: Hint: $\sum_{k=1}^{\infty} \frac{1}{k}$ and $ \sum_{k=1}^{\infty} \frac{1}{n+k+1} $ are both infinity so you cannot write these sums separately. Instead, you should write $\lim_{N \to \infty}  [-\sum_{k=1}^{N} \frac{1}{k} +\sum_{k=1}^{N} \frac{1}{n+k+1}]$.
Now $[-\sum_{k=1}^{N} \frac{1}{k} +\sum_{k=1}^{N} \frac{1}{n+k+1}]$ simplifies to $\frac 1 {N+1}+\frac 1 {N+2}+..+\frac 1 {n+N+1}-(1+\frac 1  2+\frac 1  3+...+\frac 1 {n+1})$ (for $N>n+1$). Note that $\frac 1 {N+1}+\frac 1 {N+2}+..+\frac 1 {n+N+1} \to 0$ as $N \to \infty$.
A: Just for your curiosity
Assuming that you enjoy the gaussian hypergeometric function, there is an antiderivative
$$I_n(x)=\int \left (1-\frac{x}{n} \right ) ^n \log(x)\, dx=\frac 1{n^n}\int(n-x)^n \log(x)\,dx $$
$$I_n(x)=-\frac{(n-x)^{n+1}}{n^{n+1} (n+1) (n+2)}\Big[(n-x) \, _2F_1\left(1,n+2;n+3;\frac{n-x}{n}\right)+n (n+2) \log (x) \Big]$$
$I_n(n)=0$ and then
$$\int_0^n \left (1-\frac{x}{n} \right ) ^n \log(x)\, dx=\frac{n (\log (n)-\psi (n+2)-\gamma )}{n+1}=\frac{n }{n+1}\left(\log (n)-H_{n+1}\right)$$
