Question: The equation is $$y'' + y=p(x)$$ Let $p(x)=\sum\limits\_{k=0}^n a\_k x^k$  

The homogeneous solution to the equation is $y\_h = a\cos(x)+b\sin(x)$. I can look for a particular solution of the form $y\_p=\sum\limits\_{k=0}^n b\_k x^k$. 
We get $$\sum\limits\_{k=0}^{n-2} ((k+2)(k+1)b\_{k+2}+b\_k)x^k+b\_{n-1}x^{n-1}+b\_nx^n=\sum\limits\_{k=0}^n a\_kx^k$$ Hence, if $0\le k \le n-2$, $(k+2)(k+1)b\_{k+2}+b\_k=a\_k$ and $b\_{n-1}=a\_{n-1}$, $b\_n=a\_n$  

I'm trying to show that $y\_p=\sum\limits\_{n=0}^{\infty}(-1)^n p^{(2n)}(x)$ from here, how do I proceed?

Answer: 
Use $t = \log^{n+1}(x) \implies \frac{1}{n+1}e^{t^{\frac{1}{n+1}}}dt = \log^n(x)dx$:

$$I = \lim\_{n\to\infty} \frac{n}{n+1}\int\_{\log^{n+1}(2)}^1 e^{t^{\frac{1}{n+1}}}\:dt \longrightarrow \int\_0^1 e\:dt = e$$

by dominated convergence.
Fixed $\epsilon\in(0,1-\ln2)$. Let $u=\ln x$ and then
$$ n\int\_2^e (\ln x)^n dx=n \int\_{\ln 2}^1 u^ne^u du=n \int\_{\ln 2}^{1-\epsilon} u^ne^u du+n \int^1\_{1-\epsilon} u^ne^u du.$$
For the 1st part,
$$ 0\le n \int\_{\ln 2}^{1-\epsilon} u^ne^u du\le 3n(1-\epsilon)^n $$
and one has
$$ \lim\_{n\to\infty} n \int\_{\ln 2}^{1-\epsilon} u^ne^u du=0. $$
For the 1st part, by the Mean Value Theorem for integrals, there is $\xi\in(1-\epsilon,1)$ such that
$$ n \int\_{1-\epsilon}^1 u^ne^u du=n e^\xi\int\_{1-\epsilon}^1 u^n du=\frac{n}{n+1}e^\xi(1-(1-\epsilon)^{n+1}). $$
Noting that
$$ e^{1-\epsilon}(1-(1-\epsilon)^{n+1})\le e^\xi(1-(1-\epsilon)^{n+1})\le e $$
one has
$$ \frac{n}{n+1}e^{1-\epsilon}(1-(1-\epsilon)^{n+1})\le n \int\_{1-\epsilon}^1 u^ne^u du=n e^\xi\int\_{1-\epsilon}^1 u^n du\le\frac{n}{n+1}e $$
and hence
$$ e^{1-\epsilon}\le \underline{\lim}\_{n\to\infty}n \int\_{1-\epsilon}^1 u^ne^u du\le \overline{\lim}\_{n\to\infty}n \int\_{1-\epsilon}^1 u^ne^u du\le e. $$
So
$$ e^{1-\epsilon}\le \underline{\lim}\_{n\to\infty}n \int\_{\ln 2}^1 u^ne^u du\le \overline{\lim}\_{n\to\infty}n \int\_{\ln 2}^1 u^ne^u du\le e. $$
Letting $\epsilon\to0$ one has
$$ e\le \underline{\lim}\_{n\to\infty}n \int\_{\ln 2}^1 u^ne^u du\le \overline{\lim}\_{n\to\infty}n \int\_{\ln 2}^1 u^ne^u du\le e $$
or
$$ \lim\_{n\to\infty}n \int\_{\ln 2}^1 u^ne^u du= e $$