$\lim_\limits{n\to \infty}\ (\frac{1}{2n} + \frac{1}{2n-1} + \frac{1}{2n-2} + ... + \frac{1}{n+1})$? what is the value given to this limit?
$$\lim_{n\to \infty}\ \bigg(\frac{1}{2n} + \frac{1}{2n-1} + \frac{1}{2n-2} + \cdots + \frac{1}{n+1}\bigg)$$
Is it simply 0 because each term tends to 0 and you are just summing up zeros? 
 A: 
I thought it might be instructive to present an approach that relies on developing an alternative representation of the series of interest.  To that end we proceed.


Note that we can write the sum $\sum_{k=1}^{n}\frac{1}{k+n}=\frac{1}{2n}+\frac{1}{2n-1}+\cdots \frac{1}{n+1}$ as
$$\begin{align}
\sum_{k=1}^{n}\frac{1}{k+n}&=\sum_{k=n+1}^{2n}\frac1k\\\\
&=\color{blue}{\sum_{k=1}^{2n}\frac1k} - \color{red}{\sum_{k=1}^n\frac1k}\\\\
&=\color{blue}{\sum_{k=1}^n\left(\frac{1}{2k-1}+\frac{1}{2k}\right)}-\color{red}{ 2\sum_{k=1}^n\frac1{2k}}\\\\
&=\sum_{k=1}^n\left(\frac{1}{2k-1}-\frac{1}{2k}\right)\\\\
&=\sum_{k=1}^n \frac{(-1)^{k-1}}{k}
\end{align}$$

Therefore, the series of interest is the alternating harmonic series.


Next, recalling that the Taylor Series for $\log(1+x)$ is $\log(1+x)=\sum_{k=1}^\infty \frac{(-1)^{k-1}x^k}{k}$, we have $\log(2)=\sum_{k=1}^n \frac{(-1)^{k-1}}{k}$.  Therefore, we assert that

$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to \infty} \sum_{k=1}^{n}\frac{1}{k+n}=\log(2)}$$



NOTE: Evaluating the alternating harmonic series without appealing to Taylor Series

We can evaluate the alternating harmonic series a number of ways.  Here we proceed by writing
$$\begin{align}
\sum_{k=1}^\infty \frac{(-1)^{k-1}}{k}&=\sum_{k=1}^\infty (-1)^{k-1}\int_0^1 x^{k-1}\,dx\\\\
&=\lim_{n\to \infty}\int_0^1 \sum_{k=1}^n (-x)^{k-1}\,dx\\\\
&=\lim_{n\to \infty}\int_0^1\frac{1-(-x)^n}{1+x}\,dx\\\\
&=\log(2)-\lim_{n\to \infty}\int_0^1 \frac{(-x)^n}{1+x}\,dx
\end{align}$$
For the integral on the right-hand side of the last expression, the even terms monotonically decrease while the odd terms monotonically increase in $n$.  Inasmuch as the absolute value of the integral is bounded as
$$\left|\int_0^1 \frac{(-x)^n}{1+x}\,dx \right|\le \frac{1}{n+1}$$
the limit $\lim_{n\to \infty}\int_0^1 \frac{(-x)^n}{1+x}\,dx$ exists and is equal to $0$.  In fact, using Leibniz's test, the alternating series converges and therefore, the limit $\lim_{n\to \infty}\int_0^1 \frac{(-x)^n}{1+x}\,dx$ must exist.
Therefore, we arrive at the value of the series of interest as $\log(2)$ as expected!
A: Note that each term tends to $0$ but the number of terms tends to $\infty$.
Hint. On may write
$$
\frac{1}{2n} + \frac{1}{2n-1} + \frac{1}{2n-2} + ... + \frac{1}{n+1}=\frac1n \cdot \sum_{k=1}^n\frac1{1+\frac{k}{n}}
$$ and recognize a Riemann sum.
A: By using this, 
we get $\lim_{n\to \infty}(\frac{1}{2n} + \frac{1}{2n-1} + \frac{1}{2n-2} + ... + \frac{1}{n+1})=\ln{2}$.
A: For $n>0\;$ and $\;k=n+1,n+2,...2n,$
$$\int_k^{k+1}\frac{dx}{x}\leq \frac{1}{k}\leq \int_{k-1}^k \frac{dx}{x}$$
$\implies$
$$\ln(\frac{2n+1}{n+1})\leq\sum_{n+1}^{2n}\frac 1k\leq \ln(\frac{2n}{n})$$
and the limit is $\ln(2)$.
A: Alternatively,
\begin{align*}
  1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n} &=
  \ln n+\gamma+\frac{1}{2n}+O\left( \frac{1}{n^2} \right) \\
  \frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{2n} &=
  \left[ \ln (2n)+\gamma+\frac{1}{2(2n)} \right]-
  \left( \ln n+\gamma+\frac{1}{2n} \right)+
  O\left( \frac{1}{n^2} \right) \\
  &= \ln 2-\frac{1}{4n}+O\left( \frac{1}{n^2} \right)
\end{align*}
A: More generally,
using
$\frac1{k}
\gt \int_k^{k+1} \frac{dt}{t}
\gt \frac1{k+1}
$,
we have
$\sum_{an}^{bn} \frac1{k}
\gt \sum_{an}^{bn} \int_k^{k+1} \frac{dt}{t}
= \int_{an}^{bn+1} \frac{dt}{t}
=\int_{an}^{bn} \frac{dt}{t}+\int_{bn}^{bn+1} \frac{dt}{t}
$
so
$\sum_{an}^{bn} \frac1{k}-\int_{an}^{bn} \frac{dt}{t}
\gt \int_{bn}^{bn+1} \frac{dt}{t}
\gt \frac1{bn}
$
and
$\sum_{an}^{bn} \frac1{k}
\lt \sum_{an}^{bn} \int_{k-1}^{k} \frac{dt}{t}
= \int_{an-1}^{bn} \frac{dt}{t}
=\int_{an}^{bn} \frac{dt}{t}+\int_{an-1}^{a} \frac{dt}{t}
$
so
$\sum_{an}^{bn} \frac1{k}-\int_{an}^{bn} \frac{dt}{t}
\lt \int_{an-1}^{an} \frac{dt}{t}
\lt \frac1{an-1}
$.
Since
$\int_{an}^{bn} \frac{dt}{t}
=\ln(bn)-\ln(an)
=\ln\frac{b}{a}
$,
$\frac1{bn}
\lt \sum_{an}^{bn} \frac1{k}-\ln\frac{b}{a}
\lt \frac1{an-1}
$
so
$\lim_{n \to \infty} \sum_{an}^{bn} \frac1{k}
=\ln\frac{b}{a}
$.
Note that
any fixed number of terms
added or subtracted 
from the ends of
$\sum_{an}^{bn} \frac1{k}
$
will not change the limit
since the sum
will be changed by
at most
$\frac{c}{n}$
where $c$ depends on
$a, b,$
and the number of terms.
