Let $s_{n} := 1 + 1/2 + \cdots + 1/n - \ln(n+1)$ and show that $s_{n} \leq 1 - 1/(n+1)$.

Question

This is in a Riemann integration section of a practice sheet from real analysis. I know that if $f(x) := \frac{1}{[x]} - \frac{1}{x+1}$, where $[*]$ is the floor function, then $\int_{1}^{n} f = s_{n}$. But I'm not sure where to go to get the quantity on the right-hand side of the inequality. I feel that I should compare $f \leq g$ for some $g$ where $\int_{1}^{n} g = 1- \frac{1}{n+1}$, but I am having trouble doing that!

Answer 1

Induction does the trick

Assume it true for $n-1$, i.e. $$1 + \ldots + \frac{1}{n-1} - \ln n < 1 - \frac{1}{n}$$ Then $$1 + \ldots + \frac{1}{n-1} + \frac{1}{n} - \ln (n+1) < \ln n + 1 - \frac{1}{n} + \frac{1}{n} - \ln (n+1) = 1 + \ln (\frac{n}{n+1}) = 1 + \ln(1 - \frac{1}{n+1})$$ Note that $$\ln (1-x) < -x$$ for $0 < x < 1$, hence $$ 1 + \ln(1 - \frac{1}{n+1}) < 1 - \frac{1}{n+1}$$ We conclude $$1 + \ldots + \frac{1}{n-1} + \frac{1}{n} - \ln (n+1) < 1 - \frac{1}{n+1}$$

Answer 2

If you want to use Riemann integration, then I suggest you integrate the funcion $1/x$ on the interval $[1, n+1]$. On the one hand, it is $\ln(n+1)$, on the other hand, $1+\cdots +1/n$ is clearly an upper estimation for it. In fact, if you partition the interval into $n$ subintervals of length 1, produce the rectangles for the upper Riemann sum, and cancel the area from those that are below the graph of $1/x$, then you can shift all these irregular areas into a unit square, but all those little parts are above the horizontal line that intersects the $y$ axis at $1/(n+1)$. So the "surplus" is at most the area of the rectangle with sides $1$ and $1-1/(n+1)$.

Answer 3

We have

$$s_n=\sum_{k=1}^{n}\frac1k-\ln(n+1)=\sum_{k=1}^{n}\left(\frac1k-\int_k^{k+1}\frac{dt}{t}\right)\le \sum_{k=1}^n\left(\frac1k-\frac1{k+1}\right)=1-\frac1{1+n}$$ where the last equality is obtained by telescoping.