Prove summation inequality with induction

Question

Yesterday at university a professor gave us two problems that left many doubts.

1) $\displaystyle \sum_{i=1}^n \frac{1}{i^2} \leq 2-\frac1{n}$,

2) $\displaystyle \sum_{i=1}^n \frac1{n+i} \leq \frac3{4}$.

I tried to solve both using the "classic" induction way initially, so I try for $n=0$ first, then I assume $n$ is right and finally I try with $n+1$; with some shift in the summation etc, in both situation I could say that the second term of "$n+1$" is always bigger than the second term of "$n$", so if I say so, is this enough? Are there other way to solve this? Thanks a lot!

Answer 1

As regards the first sum, we have that $$\sum_{i=1}^n \frac{1}{i^2}\leq 1+\sum_{i=2}^n \frac{1}{i(i-1)}= 1+\sum_{i=2}^n \left(\frac{1}{i-1}-\frac{1}{i}\right)=1+1-\frac{1}{n}=2-\frac{1}{n}.$$

Hint for the second sum. Show that for $i=1,\dots,n$, $$\frac1{n+i}+\frac1{n+(n+1-i)}\leq \frac{3}{2n}.$$

Answer 2

For the first, one can establish

$$S_{n+1}=S_n+\frac1{(n+1)^2}\le 2-\frac1n+\frac1{(n+1)^2}\le2-\frac1{n+1},$$

which is true because $-(n+1)^2+n\le-n(n+1)$.

For the second,

$$S_{n+1}=S_n-\frac1{n+1}+\frac1{2n+1}\le\frac34-\frac1{n+1}+\frac1{2n+1}\le\frac34$$ because $n+1<2n+1$.