Partial reciprocal sum How can I show that $$\sum_{k=1}^{n}\frac{1}{n+k}\leq\frac{3}{4}$$ for every integer $n \geq 1$?
I tried induction, estimates with logarithms and trying to bound the sum focusing on the larger terms or things like $\frac{1}{n+1}+\frac{1}{n+2}\leq\frac{2}{n+1}$ but nothing seems to work. Do you have any suggestion? Thanks
 A: We have
$$e^x = \lim_{n\to\infty}\left(1+\frac xn\right)^n\ge1+x$$
by Bernoulli's inequality. This implies that $e^{-x} \ge 1-x$, so
$$x \le -\ln(1-x).$$
Letting $x=\frac{1}{n+k}$ forms
$$\frac{1}{n+k}\le -\ln\left(\frac{n+k-1}{n+k}\right)$$
$$\frac{1}{n+k}\le\ln(n+k)-\ln(n+k-1)$$
The sum on the right telescopes giving
$$\sum_{k=1}^{n}\frac{1}{n+k}\le \sum_{k=1}^{n}\ln(n+k)-\ln(n+k-1)=\ln(2n)-\ln(n)=\ln(2),$$
by which $\ln(2)<{3}/{4}$ and we are done.
A: Here is an elementary proof. Let $$ u_n = \sum_{k=1}^n \frac{1}{n+k}$$
For all $n \geq 1$, we have
$$u_{n+1}-u_n = \sum_{k=1}^{n+1} \frac{1}{n+1+k} - \sum_{k=1}^n \frac{1}{n+k}$$ $$= \sum_{k=2}^{n+2} \frac{1}{n+k} - \sum_{k=1}^n \frac{1}{n+k} = \frac{1}{2n+2} + \frac{1}{2n+1} - \frac{1}{n+1}$$ $$ = \frac{1}{2n+1}-\frac{1}{2n+2} = \frac{1}{(2n+1)(2n+2)} > 0.$$
So the sequence $(u_n)_{n \geq 1}$ is strictly increasing.
Moreover, you have
$$u_n = \frac{1}{n} \sum_{k=1}^n \frac{1}{1 + \frac{k}{n}} \longrightarrow \int_0^1 \frac{1}{1+x} \mathrm{dx} = \ln(2)$$
So the sequence is increasing and converges to $\ln(2)$, so you have $u_n \leq \ln(2)$ for all $n \geq 1$. You deduce
$$u_n \leq \frac{3}{4}$$
A: This sum $$\sum_{k=1}^{n}\frac{1}{n+k}$$ can be converted into $$\frac{1}{n}\sum_{k=1}^{n}\frac{1}{1+\frac{k}{n}}$$
When $n$ is a large number, we get $$\int_{0}^{1}\frac{1}{1+x}dx=\ln(2)$$
A graph for reference:https://www.desmos.com/calculator/7titqyfqbx
Since $\ln(2)$ is smaller than $0.75$ the inequality is hence proved.
A: $s(n)
=\sum_{k=1}^n \dfrac1{k+n}
=\dfrac1{n}\sum_{k=1}^n \dfrac1{1+k/n}
$.
I will show that
$0
\lt \ln(2)-s(n)
\lt \dfrac1{2n+1}
$
so
$\ln(2)-\dfrac1{2n+1}
\lt s(n)
\lt \ln(2)
$.
$\dfrac1{1+k/n}
\le n\int_{(k-1)/n}^{k/n} \dfrac{dt}{1+t}
= n(\ln(1+k/n)-\ln(1+(k-1)/n))
$
so
$s(n)
\lt\dfrac1{n}\sum_{k=1}^n n(\ln(1+k/n)-\ln(1+(k-1)/n))
=\ln(2)-\ln(1)
=\ln(2)
$.
Also
$\dfrac1{1+k/n}
\gt n\int_{k/n}^{(k+1)/n} \dfrac{dt}{1+t}
= n(\ln(1+(k+1)/n)-\ln(1+k/n))
$
so
$\begin{array}\\
s(n)
&\gt\dfrac1{n}\sum_{k=1}^n n(\ln(1+(k+1)/n)-\ln(1+k/n))\\
&=\ln(2+1/n)-\ln(1+1/n)\\
&=\ln(2)+\ln(1+1/(2n))-\ln(1+1/n)\\
&=\ln(2)+\ln\left(\dfrac{1+1/(2n)}{1+1/n}\right)\\
&=\ln(2)+\ln\left(\dfrac{2n+1}{2n+2}\right)\\
&=\ln(2)-\ln\left(1+\dfrac{1}{2n+1}\right)\\
&\gt\ln(2)-\dfrac{1}{2n+1}
\qquad\text{since }\ln(1+x) < x\\
\end{array}
$
A: First, note that $f(n)=\sum_{k=1}^n\frac{1}{k+n}$.  When $n$ is a positive integer greater than 1, this is a special case of $g(n)=\int_{n+1}^{2n+1}\frac{1}{\lfloor x\rfloor}dx$
$$\frac{dg}{dn}=\frac{d}{dn}\int_{n+1}^{2n+1}\frac{1}{\lfloor x\rfloor}dx=2\frac{1}{\lfloor 2n+1\rfloor}-\frac{1}{\lfloor n+1\rfloor}=\frac{2\lfloor n+1\rfloor-\lfloor2n+1\rfloor}{\lfloor1+2n\rfloor\lfloor1+n\rfloor}$$
For positive $n$, the numerator of this is $1$ if the fractional part of $n$ less than $\frac{1}{2}$, and $0$ is the fractional part is greater than $\frac{1}{2}$, and the denominator is positive.  As such, $g(n)$ is an increasing function for all positive $n$.
If a function $g(n)$ is increasing over the domain $(1,\infty)$, then it has an upper bound of $\lim_{n\rightarrow\infty}g(n)$
$$\lim_{n\rightarrow\infty}g(n)=\lim_{n\rightarrow\infty}\int_{n+1}^{2n+1}\frac{1}{\lfloor x\rfloor}dx=\lim_{n\rightarrow\infty}\int_{1}^{2n+1}\frac{1}{\lfloor x\rfloor}dx-\int_{1}^{n+1}\frac{1}{\lfloor x\rfloor}dx+\ln(2n+1)-\int_1^{2n+1}\frac{1}{x}dx-\ln(n+1)+\int_1^{n+1}\frac{1}{x}dx=\lim_{n\rightarrow\infty}\int_{1}^{2n+1}\frac{1}{\lfloor x\rfloor}-\frac{1}{x}dx-\int_{1}^{n+1}\frac{1}{\lfloor x\rfloor}-\frac{1}{2}dx+\ln(\frac{2n+1}{n+1})=\lim_{n\rightarrow\infty}\int_{1}^{2n+1}\frac{1}{\lfloor x\rfloor}-\frac{1}{x}dx-\lim_{n\rightarrow\infty}\int_{1}^{n+1}\frac{1}{\lfloor x\rfloor}-\frac{1}{2}dx+\lim_{n\rightarrow\infty}\ln(\frac{2n+1}{n+1})=\gamma-\gamma+\ln(2)=\ln(2)$$
Here, $\gamma$ is the Euler Mascheroni constant, defined as $\lim_{m\rightarrow\infty}\sum_{i=1}^m\frac{1}{i}-\ln(m)$.
We found that $g(n)$, and therefore our sum, has an upper bound of $\ln(2)$.  As $\ln(2)\approx0.69<\frac{3}{4}$, we have proven our sum is less than $\frac{3}{4}$ for all $n>1$
