Limit of a real sequence Let
$$x_{n}=\frac{1}{n+2^{0}}+\frac{1}{n+2^{1}}+...+\frac{1}{n+2^{n}}\quad (n\in\mathbb{N}, \text{ }n\geq 1).$$
What is the limit of $x_n$?
Attempt:  Just the boundedness, $x_n∈(0,2)$, $\forall n\geq 1$. Can not even determine the monotonicity.
 A: $\begin{array}\\
x_n
&=\sum_{k=0}^n \dfrac1{n+2^k}\\
&=\sum_{k=0}^{\log_2(n)} \dfrac1{n+2^k}+\sum_{k>\log_2(n)}^n \dfrac1{n+2^k}\\
&\lt\sum_{k=0}^{\log_2(n)} \dfrac1{n}+\sum_{k>\log_2(n)}^n \dfrac1{2\cdot 2^k}\\
&\lt \dfrac{\log_2(n)+1}{n}+\dfrac1{n}\\
&\to 0\\
\end{array}
$
A: Just for your curiosity.
Since @marty cohen already provided the good solution and the upper bound, I shall just report here some results we obtained in my former research group a few decades ago working the most general problem
$$x_n^{(a)}=\sum_{k=0}^n \dfrac1{n+a^k}\qquad \text{where} \qquad a >1$$
The general formula is given by
$$x_n^{(a)}=\frac{\psi _{\frac{1}{a}}^{(0)}\left(-\frac{\log (-n)}{\log (a)}\right)-\psi
   _{\frac{1}{a}}^{(0)}\left(n+1-\frac{\log (-n)}{\log (a)}\right)}{n \log (a)}$$ where appears the q-polygamma function which is related to Lambert series.
As good approximation was proposed (from some deep arguments)
$$x_n^{(a)}=\frac {\alpha_a+(1-\beta_a) \log(n)}{n \log(a)}$$ where $\alpha_a <1$ increases and $\beta_a$ is a small number which decrease when $a$ increases.
For the specific case of $a=2$, $\alpha_2 \sim 0.3877$ and $\beta_2\sim 0.0063$.
