Prove that $\sum_{k=1}^{n} \frac{1}{n+k} < \ln 2$ I have a loose upper bound:
$$
\sum_{k=1}^{n} \frac{1}{n+k}  < \frac{1}{n} \cdot  n = 1
$$
Clearly this is very loose compared to the upper bound $\ln 2 \approx 0.693$
 A: 
I thought it might be instructive to present an approach that does not rely on Riemann sums or the asymptotic series for the harmonic series.  Rather, we use only the Taylor series for the logarithm function and straightforward arithmetic.  To the end, we begin with a primer.


PRIMER:   Taylor Series for $\displaystyle \log(1+x)$ 
The Taylor series for $\log(1+x)$ is given by
$$\log(1+x)=\sum_{k=1}^\infty\frac{(-1)^{k-1}x^k}{k} \tag1$$
for $-1<x\le 1$.  Setting $x=1$ in $(1)$ reveals 
$$\log(2)=\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k}\tag2$$
We will use the result in $(2)$ in the following development. 

Note that we can write
$$\begin{align}
S_n&=\sum_{k=1}^n\frac{1}{n+k}\\\\
&=\sum_{k=n+1}^{2n}\frac1k\\\\
&=\sum_{k=1}^{2n}\frac1k-\sum_{k=1}^n\frac1k\\\\
&=\sum_{k=1}^n \left(\frac{1}{2k-1}+\frac{1}{2k}\right)-\sum_{k=1}^n \frac1k\\\\
&=\sum_{k=1}^n \left(\frac{1}{2k-1}-\frac{1}{2k}\right)\\\\
&=\sum_{k=1}^{2n}\frac{(-1)^{k-1}}{k}\tag3
\end{align}$$
Now, $S_n$ is an alternating series with $\frac1k$ monotonically decreasing to $0$.  Moreover, from $(3)$ it is easy to see that $S_{n+1}>S_n>0$.  Therefore,  using the behavior of this alternating series and applying $(2)$ yields
$$\begin{align}
S_n&\le \sum_{k=1}^\infty \frac{(-1)^{k-1}}{k}\\\\
&=\log(2)
\end{align}$$
as was to be shown!

TOOLS USED:  Taylor series for $\log(1+x)$, observation on behavior of partial sums of alternating series, straightforward arithmetic.

A: Hint:
Rewrite the sum as $\;\displaystyle\frac1n\sum_{k=1}^n\frac1{1+\smash[b]{\cfrac kn}}$, and you fall on a lower Riemann sum for the function $\dfrac1x$ on the interval $[1,2]$.
A: We know  $H_n = \sum_{k = 1}^n\frac{1}{k} \leq \ln(n)+1$ from this link. Hence:
$$
\sum_{k=1}^{n} \frac{1}{n+k} =  H_{2n}-H_n \leq \ln(2n)-\ln(n)= \ln(2)
$$
A: Just for fun here is an alternative almost correct(?) solution:
Let
$$\sum_{k=1}^{n}\frac{1}{n+k}=\sum a_k$$
$$e^{\sum a_k}=e^{a_1}e^{a_2}...e^{a_n}\leq\left(1+\frac1n\right)\left(1+\frac1{n+1}\right)...\left(1+\frac1{2n-1}\right)=1+\frac1n+\frac1{n+1}+...+\frac1{2n-1}+o\left(\frac1{n^2}\right)\leq1+n\frac1n+o\left(\frac1{n^2}\right)=2+o\left(\frac1{n^2}\right)$$
thus
$$\sum_{k=1}^{n}\frac{1}{n+k}=\log e^{\sum a_k}\leq \log \left[2+o\left(\frac1{n^2}\right)\right]=\log 2 +o\left(\frac1{n^2}\right)$$
