Doubt on limit of sum of sequences. Two procedures leads to different answers. I have the following problem.
Determine the convergence or divergence of the sequence $(x_n)$ where $$x_n=\frac{1}{n+1} + \frac{1}{n+2} + \cdots + \frac{1}{2n}.$$
My first approach was: 
Well, since $(x_n)$ is the sum of the sequences $\left(\frac{1}{n+1}\right),\ldots,\left(\frac{1}{2n}\right)$ and each of them are convergent, in fact they converges to $0$, then the limit of $(x_n)$ must be $0$.
My second approach was: 
If we make the calculations, we have that $$x_{n+1}-x_n=\frac{1}{(2n+1)(2n+2)}>0$$ then $$x_{n+1}>x_n$$ i.e., $x_n$ is an increasing sequence.
Also, as in $x_n$ we sum $n$ elements which are less or equan than $\frac{1}{n+1}$,  then $$x_n=\frac{1}{n+1} + \frac{1}{n+2} + \ldots + \frac{1}{2n}<\frac{n}{n+1}<1$$ i.e., $(x_n)$ is a bounded sequence.
Now, since $(x_n)$ is increasing and bounded sequence, it follows that its limit is the supremum of $(x_n)$. 
Annoying question.
But how $(x_n)$ can converge to $0$ (as shown in my first approach) and at the same time converges to its supremum when it is an increasing sequence? Can be 0 its supremum? It is not a contradiction? 
Any help would be appreciated.
 A: You can use a well-known result in analytic number theory that $H_n = 1+\dfrac{1}{2}+\dfrac{1}{3} +\cdots +\dfrac{1}{n} - \ln n$ converges to $\gamma$, and write $x_n$ in terms of $H_n$. Or you can use Riemann sum as a preferred method as follows: $x_n \to \displaystyle \int_{0}^1 \dfrac{1}{1+x}dx = \ln 2$
A: I thought it might be instructive to present an approach that relies on only straightforward arithmetic and Leibniz's Test for alternating series.  
Proceeding with this way forward we first rewrite the series of interest $\sum_{k=n+1}^{2n}\frac{1}{k}$ as
$$\begin{align}
\sum_{k=n+1}^{2n}\frac{1}{k}&=\sum_{k=1}^{2n}\frac{1}{k}-\sum_{k=1}^{n}\frac{1}{k}\\\\
&=\sum_{k=1}^{n}\left(\frac{1}{2k-1}+\frac{1}{2k}\right)-2\sum_{k=1}^{n}\frac{1}{2k}\\\\
&=\sum_{k=1}^{n}\left(\frac{1}{2k-1}-\frac{1}{2k}\right)\\\\
&=\bbox[5px,border:2px solid #C0A000]{\sum_{k=1}^{2n}(-1)^{n-1}\frac1k} \tag 1
\end{align}$$
Then, since $\frac1k$ converges monotonically to zero, Leibniz's Test guarantees that the series in $(1)$ converges.  

It might be interesting to note that while the series on the right-hand side of $(1)$ converges conditionally, the series $\sum_{k=1}^\infty = \left(\frac{1}{2k-1}-\frac{1}{2k}\right)=\sum_{k=1}^\infty \frac{1}{2k(2k-1)}$ converges absolutely.  Therefore, we have two different convergent series representations, only one of which is absolutely convergent.

Note, we can evaluate the limit of series on the right-hand side of $(1)$ by recalling the Taylor series for $\log(1+x)$ is given by 
$$\log(1+x)=\sum_{k=1}^{\infty}(-1)^{n-1}\frac{x^k}{k}\tag 2$$ 
Comparing $(1)$ and $(2)$ we see that the limit of the series of interest is given by
$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to \infty}\sum_{k=n+1}^{2n}\frac{1}{k}=\log(2)}$$ 
which agrees with the results reported that used Riemann sums!
A: Proposed idea: Each of the terms converges to $0$; therefore their sum converges to $0$. [???]
That is true if the number of terms doesn't change as $n$ grows.  If $A_1,\ldots,A_{50}$ all approach $0$ as $n\to\infty$ then so does $A_1+\cdots+A_{50}$.  But if instead of $50$ terms, we have $n$ terms, so the number of terms grows as $n$ grows, then the conclusion is not longer valid.  For example consider this:
$$
\underbrace{ \frac 1 n + \cdots + \frac 1 n }_\text{$n$ terms}.
$$
This adds up to $1$, and that doesn't approach $0$ as $n$ grows, even though every term approaches $0$ as $n$ grows.
This sum can be viewed as a Riemann sum:
$$
\frac 1 n \left( f(1) + f\left(1+\frac 1 n\right) + f\left(1+\frac 2n\right) + \cdots + f\left( 1+ \frac n n \right) \right)
$$
where $f(x) = \dfrac 1 x$.  So this sum approaches $\displaystyle \int_1^2 \frac 1 x\,dx = \log_e 2$.
A: If you use harmonic numbers, you have $$x_n=H_{2 n}-H_n$$  For large values of $k$, the Wikipedia page gives $$H_k=\gamma +\log (k)+\frac{1}{2 k}-\frac{1}{12
   k^2}+O\left(\frac{1}{k^3}\right)$$ which makes $$x_n=\log (2)-\frac{1}{4 n}+\frac{1}{16 n^2}+O\left(\frac{1}{n^3}\right)$$ which shows the limit and also how it is approached.
For illustration purposes, let us us $n=10$; the exact result is $\frac{155685007}{232792560}\approx 0.66877140$ while the above approximation would give $\log(2)-\frac{39}{1600}\approx 0.66877218$
