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 :

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


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

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.

  • 2
    $\begingroup$ 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. $\endgroup$ – 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).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.