Doing proof by induction exercises with inequalities, I got stuck on one that is a bit different from the others. There is no $n$ term on the rightmost part of the inequality:
Prove that the following holds for all $n \ge 1$:
$$\frac{1}{n + 1}+\frac{1}{n + 2}+...+\frac{1}{n+(n+1)}\le \frac{5}{6}$$
All the other proofs I did before had some $n$ involved there, but now there is none. I wonder how will this difference affect my standard attempt:
Test for $n = 1$:
$$\frac{1}{3} \le \frac{5}{6} \implies 6 \le 15$$
We have to prove that it holds for $n + 1$, that is:
$$\frac{1}{n + 1}+\frac{1}{n + 2}+...+\frac{1}{n+(n+1)}+\frac{1}{(n + 1)+(n+2)}\le \frac{5}{6}$$
Asssume $$\frac{1}{n + 1}+\frac{1}{n + 2}+...+\frac{1}{n+(n+1)}\le \frac{5}{6}$$
We start off with
$$\frac{1}{n + 1}+\frac{1}{n + 2}+...+\frac{1}{n+(n+1)}+\frac{1}{(n + 1)+(n+2)}\le \frac{5}{6}$$
And now I see the problem. Normally, I apply the hypothesis first on this part:
$$\color{blue}{\frac{1}{n + 1}+\frac{1}{n + 2}+...+\frac{1}{n+(n+1)}}+\frac{1}{(n + 1)+(n+2)}\le \frac{5}{6}$$
But all it yields is this:
$$\color{blue}{\left ( \textrm{something smaller or equal than } \frac{5}{6} \right )} + \frac{1}{(n + 1)+(n+2)}\le \frac{5}{6}$$
If I move $\frac{1}{(n + 1)+(n+2)}$ to the other side, all I get is
$$\color{blue}{\left ( \textrm{something smaller or equal than } \frac{5}{6} \right )} \le \frac{5}{6} - \frac{1}{(n + 1)+(n+2)}$$
Which cannot be guaranteed.
How can I prove this, then?
Let $$a_n=\sum_{k=n+1}^{2n+1}\frac1k\;,$$ and show that the sequence $\langle a_n:n\ge 1\rangle$ is decreasing. There’s more in the spoiler-protected block below.