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: 
As you said $\lim\limits\_{n\to \infty}\ln^n 2=0$. Let:

$$I\_n = \int\_2^e\ln^n x\,dx$$

$I\_n$ is decreasing because:

$$I\_{n}-I\_{n+1}=\int\_2^e\ln^n x(1-\ln x)\,dx \geq 0$$

Integrating by parts:

$$
\begin{aligned} 
\displaystyle\int\_2^e \ln^n x\, dx &= \displaystyle\int\_2^e \ln^n x \cdot (x)'\, dx\\
&=x\ln^n x\bigg|\_2^e -n\int\_2^e\ln^{n-1}x\,dx\\
&= e-2\ln^n 2-nI\_{n-1}
\end{aligned}
$$

Therefore $I\_n+nI\_{n-1}=e-2\ln^n2$. Now, since $I\_n$ is decreasing:

$$I\_n+nI\_{n-1}\geq I\_n+nI\_n\Rightarrow I\_n\leq \frac{1}{n+1}(e-2\ln^n 2)$$

and

$$I\_{n+1}+(n+1)I\_n\leq I\_n+(n+1)I\_n\Rightarrow I\_n\geq \frac{1}{n+2}(e-2\ln^{n+1}2)$$ 

Combining the two:

$$\frac{1}{n+2}(e-2\ln^{n+1}2) \leq I\_n\leq \frac{1}{n+1}(e-2\ln^n 2)$$

or

$$\frac{n}{n+2}(e-2\ln^{n+1}2) \leq nI\_n\leq \frac{n}{n+1}(e-2\ln^n 2)$$

and squeezing $\lim\limits\_{n\to \infty}nI\_n=e$.
Let $\epsilon $ be any fixed number such that $0<\epsilon <e-2$ and we split the integral as a sum of two integrals using intervals $[2,e-\epsilon]$ and $[e-\epsilon, e] $. The first integral tends to $0$ as integrand uniformly tends to $0$.

For the second integral we need to bound it suitably. We have via integration by parts $$(n+1)\int\_{e-\epsilon} ^{e} \dfrac{(\log x) ^n} {x} \cdot x\, dx=\left. x(\log x) ^{n+1}\right|\_{x=e-\epsilon} ^{x=e} - \int\_{e-\epsilon} ^{e}(\log x) ^{n+1}\,dx$$ (note that the factor $n$ before integral can be safely replaced by $(n+1)$ as $n/(n+1)\to 1$). The first term on right tends to $e$ and second term is bounded in absolute value by $\epsilon$ (as integrand lies in $[0,1]$). It follows that the desired limit is $e$.