Is this inequality true for all k ? $\sum_{n=k}^{n=+\infty} \frac{1}{n^4} \leq (\sum_{n=k}^{n=+\infty} \frac{1}{n^2})^3$

Can it be generalized for other powers ? Wolfram seems to say it is true for k below 20000.

I stumbled upon it randomly when trying to approximate $\sum_{n=1}^{n=+\infty} \frac{1}{n^4}$.

My reasoning was :

$$\left(\sum_{n=k}^{n=+\infty} \frac{1}{n^2}\right)^2=\sum_{n=k}^{n=+\infty} \frac{1}{n^4} + (\text{double products}) \geq\sum_{n=k}^{n=+\infty} \frac{1}{n^4}$$

So

$$\sum_{n=1}^{n=+\infty} \frac{1}{n^4} \leq \sum_{n=1}^{n=k-1} \frac{1}{n^4}+\left(\sum_{n=k}^{n=+\infty} \frac{1}{n^2}\right)^2 \leq \left(\sum_{n=1}^{n=k-1} \frac{1}{n^4}\right)+\left(\frac{1}{k-\frac{1}{2}}\right)^2$$

where the last inequality comes from An inequality: $1+\frac1{2^2}+\frac1{3^2}+\dotsb+\frac1{n^2}\lt\frac53$.

Then I noticed that, perhaps, I could raise the last term to the power of 3 instead of just 2, making the inequality stronger.

• The LHS behaves like $\int_{k}^{+\infty}\frac{dx}{x^4}=\frac{1}{3k^3}$ while the RHS behaves like $\left(\int_{k}^{+\infty}\frac{dx}{x^2}\right)^3=\frac{1}{k^3}$, so that is not surprising. – Jack D'Aurizio Apr 21 '16 at 15:36

For $k > 1$, $$\sum_{n=k}^\infty \dfrac{1}{n^4} < \int_{k-1}^\infty \dfrac{dx}{x^4} = \dfrac{1}{3(k-1)^3}$$

$$\left(\sum_{n=k}^\infty \dfrac{1}{n^2}\right)^3 > \left(\int_{k}^\infty \dfrac{dx}{x^2}\right)^3 = \dfrac{1}{k^3}$$

$\dfrac{1}{k^3} > \dfrac{1}{3(k-1)^3}$ when $3(k-1)^3 > k^3$, which is true for $k > 3.2612$.

Let's see what we can say about comparing $s_1(k) =\sum_{n=k}^{\infty} \frac1{n^a}$ with $s_2(k) =\left(\sum_{n=k}^{\infty} \frac1{n^b}\right)^c$ for large enough $k$, where $a > 1$ and $b > 1$ so the sums converge.

Using the integral approximation, $s_1(k) =\sum_{n=k}^{\infty} \frac1{n^a} \approx \int_k^{\infty} \frac{dx}{x^a} =-\frac1{(a-1)x^{a-1}}\big|_k^{\infty} =\frac1{(a-1)k^{a-1}}$.

Therefore $s_2(k) =\left(\sum_{n=k}^{\infty} \frac1{n^b}\right)^c \approx \left(\frac1{(b-1)k^{b-1}}\right)^c =\frac1{(b-1)^c k^{c(b-1)}}$, so $\dfrac{s_1(k)}{s_2(k)} \approx \dfrac{\frac1{(a-1)k^{a-1}}}{\frac1{(b-1)^c k^{c(b-1)}}} = \dfrac{{(b-1)^c k^{c(b-1)}}}{{(a-1)k^{a-1}}} = \dfrac{{(b-1)^c }}{{(a-1)}}k^{c(b-1)-(a-1)}$.

Therefore if $c(b-1) > a-1$, then $s_1(k) > s_2(k)$ for all large enough $k$; if $c(b-1) < a-1$, then $s_1(k) < s_2(k)$ for all large enough $k$;

If $c(b-1) = a-1$, then $\dfrac{s_1(k)}{s_2(k)} \approx \dfrac{{(b-1)^c }}{{(a-1)}}$, so the result depends on this ratio.

For your case of $a=4$ and $b=2$, the key difference is $c(b-1)-(a-1) =c-3$.

If $c > 3$, then $s_1(k) > s_2(k)$ for large enough $k$; if $c < 3$, then $s_1(k) < s_2(k)$ for large enough $k$.

If $c=3$, which is your case, the ratio is $\dfrac{(b-1)^c }{(a-1)} =\dfrac{1}{3} < 1$, so $s_1(k) \approx \frac13 s_2(k) < s_2(k)$ for large enough $k$, which confirm's Robert Israel's result (good thing too, because any result of mine that differs from a result of his is probably wrong).