For $n>1,$ prove that, for $\sum_{n}^{3n-1}\frac{1}{x^2} < \frac{5}{4n}$

Question

For $n>1$ prove that for $$\sum_{n}^{3n-1}\frac{1}{x^2} < \frac{5}{4n}$$

I know that $\dfrac{1}{x}>\dfrac{1}{(x+1)}$ and I'm trying to break the summation into smaller sums to work with, but I'm just not making that final bridge to the $\dfrac{5}{4n}$.

Answer 1

For $n=2$, we have:

$$\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2} < \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{3^2}+\frac{1}{3^2}=\frac{1}{4}+\frac{1}{3}=\frac{7}{12}<\frac{5}{8}$$

Now assume $n\geq 3$. Since:

$$\frac{1}{x^2} < \frac{1}{x(x-1)}=\frac{1}{x-1}-\frac{1}{x}$$

we have

$$\sum_{n}^{3n-1}\frac{1}{x^2} < \sum_{n}^{3n-1}\left(\frac{1}{x-1}-\frac{1}{x}\right)=\frac{1}{n-1}-\frac{1}{3n-1}=\frac{2n}{(n-1)(3n-1)}$$

It remains to prove:

$$\frac{2n}{(n-1)(3n-1)}< \frac{5}{4n}\Leftrightarrow 7n^2-20n+5 > 0$$

which holds when $n \geq 3$.

Answer 2

Since

$$ \frac{1} {{k^2 }} < \int\limits_{k - 1}^k {\frac{1} {{x^2 }}} dx\,\,\,\,\,\,\forall k \geqslant 2 $$ we have that $$ \sum\limits_{k = n}^{3n - 1} {\frac{1} {{k^2 }} < } \int\limits_{n - 1}^{3n - 1} {\frac{1} {{x^2 }}} dx = \left[ { - \frac{1} {x}} \right]_{n - 1}^{3n - 1} = \frac{1} {{n - 1}} - \frac{1} {{3n - 1}} $$ We have that $$ \frac{1} {{n - 1}} - \frac{1} {{3n - 1}} < \frac{5} {{4n}} $$ is true if and only if $$ \frac{{2n}} {{\left( {n - 1} \right)3n - 1}} < \frac{5} {{4n}} $$ which is true if and only if $$7n^2-20n+5>0$$. This is true for every $n \geq 3$. On the other side, a direct computation shows that the inequality is true for $n=2$.

Answer 3

The telescoping sum trick, with ${1\over k^2-k}={1\over k-1}-{1\over k}$, works, but you need to split off the first term:

$$\sum_{k=n}^{3n-1}{1\over k^2}={1\over n^2}+\sum_{k=n+1}^{3n-1}{1\over k^2}\lt{1\over n^2}+\sum_{k=n+1}^{3n-1}{1\over k^2-k}={1\over n^2}+\left({1\over n}-{1\over3n-1}\right)={5\over4n}-\left({1\over4n}+{1\over3n-1}-{1\over n^2} \right)$$

and

$${1\over4n}+{1\over3n-1}-{1\over n^2}={(3n-1)n+4n^2-4(3n-1)\over4(3n-1)n^2}={7n^2-13n+4\over4(3n-1)n^2}\gt{7n(n-2)+4\over4(3n-1)n^2}\gt0$$

for $n\ge2$.