Find the limit $\lim_{n \to \infty} \frac{a_1 + a_2 + a_3 + ... + a_n}{\ln(n)}$ where $a_{n} = \int_{1}^{e} \ln^{n}(x)dx$ First I tried to solve this problem by noticing that:
$$a_1 + a_2 + a_3 + ... + a_n = \sum_{k=1}^{n}a_{k}=\sum_{k=1}^{n}\int_{1}^{e}\ln^{k}(x)dx = \int_{1}^{e}\sum_{k=1}^{n}\ln^{k}(x)dx \\= \int_{1}^{e}\frac{\ln^{k+1}(x)-\ln(x)}{1-\ln(x)}dx,$$
but this didn't really help.
Then I arrived at the equivalent form $a_n = \int_{0}^{1}x^{n}e^{x}dx$ which yields:
$$a_1 + a_2 + a_3 + ... + a_n = \int_0^1 \frac{x^{n+1}-x}{1-x}e^xdx.$$
My last attempt was to introduce a new function $F(t) = \int_0^1x^ne^{tx}dx$.
$$F^{\prime}(t)= \frac{d}{dt}\int_0^1x^ne^{tx}dx =\int_0^1 \frac{\partial}{\partial t}x^ne^{tx}dx=\int_0^1 x^n e^{tx} xdx = \int_0^1x^{n+1}e^{tx}dx.$$
By doing IBP and simplifying I got this differential equation:
$$ F^{\prime}(t)= \frac{e^t}{t}-\frac{n+1}{t}F(t)$$ which I don't know how to solve.
 A: For every $n \geq 2$, one has
$$a_n = \int_{0}^{1}x^{n}e^{x}dx = \left[x^ne^x \right]_0^1 - \int_0^1nx^{n-1}e^x dx = e-na_{n-1}$$
Adding $na_n$ to both sides, you get $(n+1)a_n = e-n(a_{n-1}-a_n)$, which rewrites as $$a_{n-1}-a_n = \dfrac{e}{n}-a_n-\dfrac{1}{n}a_n$$
Summing this for $n=2$ to $N$, you get that for every $N \geq 2$
$$a_2-a_N = e(H_N-1)-\sum_{n=2}^N a_n - \sum_{n=2}^N \dfrac{a_n}{n} \quad \quad (*)$$
where $H_N = \displaystyle{\sum_{n=1}^N \dfrac{1}{n}}$is the partial sum of order $N$ of the harmonic series.
But one has directly $$0 \leq a_n = \int_{0}^{1}x^{n}e^{x}dx \leq e \int_0^1 x^n dx = \dfrac{e}{n+1},$$
so the LHS of $(*)$ is bounded, and the second series in the RHS of $(*)$ converges, so $(*)$ gives directly that
$$ \sum_{n=2}^N a_n \sim eH_N$$
and because $H_N \sim \ln(N)$, you get that
$$\boxed{\lim_{N \rightarrow +\infty} \dfrac{1}{\ln(N)}\displaystyle{\sum_{n=1}^N a_n} =e}$$
A: This certainly is not a super formal answer however I do believe this can be formalised. I started from Stolz–Cesàro theorem  and asked about following limit $$\lim_{n\to\infty}\frac{\displaystyle\int_{1}^e\ln^nx\ \mathrm{d}x}{\ln \big(\frac{n}{n-1}\big)}.$$  Since the integral can be evaluated as @
RamanujanXXV pointed out I looked at the asymptotic behavior of the integral  $$\displaystyle\int_{1}^e\ln^nx\ \mathrm{d}x\approx \frac{e}{n}.$$ However it is easy to see that $$\ln \Big(\frac{n}{n-1}\Big)\approx \frac{1}{n-1}.$$ So the limit is equal to $e$.
A: Let us notice that
$$a_n=\int_{0}^{1}x^n e^x\,dx = \frac{e}{n+1}-\int_{0}^{1}x^n (e-e^x)\,dx.=\frac{e}{n+1}-R_n.$$
If we prove that the remainder term $R_n$ is sufficiently small we are done. The function $f_n(x)=x^n(e-e^x)$ is non-negative over $(0,1)$ with a root with multiplicity $n$ at the origin and a simple root at $1$. By the convexity of the exponential function $\frac{e-e^x}{1-x}$ is bounded between $e-1$ and $e$ over $[0,1]$, so
$$ R_n \leq e \int_{0}^{1} x^n(1-x)\,dx = \frac{e}{(n+1)(n+2)}<\frac{e}{n^2}.$$
This immediately gives
$$ \sum_{k=1}^{n}a_k = e(H_{n+1}-1)+O(1) = e\log n+O(1). $$
A: Thank you everyone for your ingenious ideas. I came up with a fairly simple answer combining the concepts I saw presented here.
$$a_n = \int_1^{e} \ln^{n}(x)dx = \int_0^{1}x^{n}e^{x}dx$$
$$\int_0^1 x^n e^x dx \leq e \int_0^1 x^ndx = \frac{e}{n+1} \implies a_n \leq \frac{e}{n+1}$$.
$$\int_0^1x^{n}e^x dx > \int_0^1x^{n+1}e^x dx \text{ (because we are working in the interval } (0, 1))$$
$$\int_0^1x^ne^xdx > e - (n+1)\int_0^1 x^n e^x dx$$
$$(n+2)\int_0^1x^ne^xdx > e \implies a_n = \int_0^1x^n e^x > \frac{e}{n+2}$$
Finally we arrive at the double inequality:
$$\frac{e}{n+2} < a_n \leq \frac{e}{n+1}$$
Then as $n \to \infty \text{, we can use the Squeeze Theorem using the double inequality above.}$.
$$\lim_{n \to \infty} \frac{a_1 + a_2 + a_3 + ... + a_{n}}{\ln(n)} = \lim_{n \to \infty} \frac{e(H_{n} - 1)}{\ln(n)} = e \lim_{n \to \infty} \frac{H_n - 1}{\ln(n)} = e \cdot 1 = \boxed{e}$$
