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
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 $\ln(2)&lt;{3}/{4}$ and we are done.
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 $\ln(2)\approx0.69&lt;\frac{3}{4}$, we have proven our sum is less than $\frac{3}{4}$ for all $n&gt;1$