Firstly, we'll show that $f_n(x) = \sum_{k=0}^n x^n$ converges uniformly to $f(x) = \frac{1}{1-x}$ on $\mathcal I_\eta=[0,1-\eta]$, for any $\eta \in (0,1)$.
To do that, note:
(1) $\forall_{x \in \mathcal I_\eta} (f_n(x))_{n \in \mathbb N } $ is an increasing sequence.
(2) $\forall_{n\in \mathbb N} f_n $ is a continuous function and so is $f$.
(3) $\forall_{\eta \in (0,1)} \mathcal I_\eta $ is compact.
(4) $(f_n)_{n \in \mathbb N}$ converges pointwise to $f$ on $\mathcal I_\eta$, for any $\eta \in (0,1)$
So, we can apply Dini's Theorem (proof can be found on wiki).
Now, due to $ \forall_{x \in \mathcal I_\eta} \forall_{n \in \mathbb N} : f_n(x) \in [1,\frac{1}{\eta}]$ (so we're on domain where function $t \to \frac{1}{t}$ is uniformly continuous), the sequence $g_n(x) = \frac{1}{f_n(x)}$ is also uniformly convergent to $g(x) = 1-x$ on $\mathcal I_\eta$, for any $\eta \in (0,1)$.
Okay. Now, we'll show that $\forall_{\eta \in (0,1)} \forall_{\epsilon>0} \exists_{N(\epsilon) \in \mathbb N} \forall_{n > N(\epsilon)}: |\int_0^{1-\eta} g_n(x)dx - \int_0^{1-\eta}g(x)dx| < \epsilon$ .
So, pick any $\eta \in (0,1), \epsilon > 0$. We have $|\int_0^{1-\eta} g_n(x)dx - \int_0^{1-\eta}g(x)dx| = |\int_0^{1-\eta} g_n(x)-g(x)dx| \leq $
$ \leq \int_0^{1-\eta} |g_n(x)-g(x)| dx = (\star)$.
Now, using the uniform convergence of $(g_n)_{n \in \mathbb N}$, we have $\forall_{\delta > 0} \exists_{M(\delta)}\forall_{n > M(\delta)}\forall_{x\in \mathcal I_\eta} |g_n(x) - g(x)| < \delta$.
Now, choose $\delta = \frac{\epsilon}{1-\eta}$ and $N(\epsilon) = M(\delta)$ so that we have:
$(\star) < \int_0^{1-\eta} \frac{\epsilon}{1-\eta} dx = \epsilon $.
We're almost done. Now, we'll show that $\forall_{\epsilon>0}\exists_{\eta \in (0,1)}\forall_{n \in \mathbb N} : |\int_{1-\eta}^1 g_n(x)dx - \int_{1-\eta}^1 g(x)dx| < \epsilon$.
Pick any $\epsilon > 0$, we have $|\int_{1-\eta}^1 g_n(x)dx - \int_{1-\eta}^1 g(x)dx| \leq \int_{1-\eta}^1 |g_n(x) - g(x)| dx = (\star) $.
Now note, that $\forall_{n \in \mathbb N}\forall_{\eta \in (0,1)}\forall_{x \in [1-\eta,1]}|g_n(x) - g(x)| \leq 2|g_{1}(0)| = 2$ (due to it being decreasing sequence of decreasing functions), so: $(\star) \leq 2\eta$ and it's sufficient to choose $\eta = \frac{\epsilon}{2}$.
We're about to complete: Choose any (small) $\epsilon > 0$. Pick $\eta = \frac{\epsilon}{2}$. We show, there exists $N(\epsilon): \forall_{n > N(\epsilon)}$:
$|\int_0^{1} g_n(x)dx - \int_0^{1}g(x)dx| \leq |\int_0^{1-\frac{\epsilon}{2}} g_n(x)-g(x)dx | + |\int_{1-\frac{\epsilon}{2}}^{1} g_n(x)-g(x)dx | < \epsilon + \epsilon = 2\epsilon$, that means, we proved:
$\lim_{n\to \infty} \int_0^1 g_n(x)dx = \int_0^1g(x)dx$.
Plugging our values, we get:
$\lim_{n\to \infty} \int_0^1 \frac{1}{\sum_{k=0}^n x^n} dx = \int_0^1 (1-x) dx = (x-\frac{x^2}{2})|^{x=1}_{x=0} = \frac{1}{2}$
