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In the study of the following series $$ \sum_{n \geq 1} {\frac{1^2+2^2+ \cdots + n^2}{n^p}} $$ it is not hard to prove that it diverges for $p \leq 3$, since the sequence itself does not converge to 0. You can also conclude that the series converges for $p > 4$ by comparison with Riemann series. Raabe's test yields that the series diverges for p between 3 and 4. However it does not give any information for the case $p=4$.

Kamal Saleh
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ChristmasTree
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3 Answers3

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For $p=4$ each term is $\frac {n(n+1)(2n+1)}{6n^4} \gt \frac 1{3n}$ The sum of these is bounded below by $\frac 13$ of the harmonic series, so the sum diverges.

Ross Millikan
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We do not need to apply the explicit formula for $1^2+2^2+\ldots +n^2.$ Instead observe that $$n^3-(n-1)^3=3n^2-3n+1\ge 3n^2$$ Hence $$1^2+2^2+\ldots +n^2\ge {1\over 3}\left ([1^3-0^3]+[2^3-1^3]+\ldots +[n^3-(n-1)^3]\right ]={1\over 3}n^3$$ Hence the $n$th term of the series for $p=4$ is greater or equal $\displaystyle{1\over 3n}.$

Another approach is to apply the Cauchy-Schwarz inequality $$(1+2+\ldots +n)^2\le (1^2+2^2+\ldots +n^2)n$$ hence $$1^2+2^2+\ldots +n^2\ge {1\over n}{n^2(n+1)^2\over 4}\ge {n^3\over 4}$$

Ryszard Szwarc
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  • That's very creative! @RysRyszard Szwarc possible mistake in last inequality when squaring the formula for the sum of consecutive integers, should be a 4 in the denominator (?) – ChristmasTree Jan 15 '23 at 22:28
  • @ ChristmasTree Thanks. Corrected – Ryszard Szwarc Jan 16 '23 at 00:30
  • +1. We can also let $n'=n/2$ when $n$ is even, and $n'=(n+1)/2$ when $n$ is odd. Then $\sum_{j=1}^n j^2\ge \sum_{j=n'}^n j^2\ge \sum_{j=n'}^n(n')^2=(n-n'+1)(n')^2>(n/2)(n')^2\ge (n/2)^3.$ – DanielWainfleet Jan 16 '23 at 04:53
  • @DanielWainfleet Thanks for upvoting. I also thought about that way, but I do not like to deal with cases. Your way with introducing $n'$ is cute. – Ryszard Szwarc Jan 16 '23 at 05:46
  • @DanielWainfleet I got an idea how to reduce to the case of even $n$: $$4(1^2+2^2+\ldots +n^2)\ge 1^2+2^2+\ldots +(2n)^2$$ – Ryszard Szwarc Jan 16 '23 at 05:58
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We know that $\displaystyle \sum _{i=1}^{n} i^{2} \ =\ \frac{n( n+1)( 2n+1)}{6}$
So $\displaystyle \frac{\sum _{i=1}^{n} i^{2}}{n^{p}} \ =\ \frac{( n+1)( 2n+1)}{6n^{p-1}}$
Now if we take the limit $\displaystyle n\rightarrow \infty $ and apply L'Hospital's Rule, we can see that
$\displaystyle \frac{4n+3}{6( p-1) n^{p-2}} \ =\ \frac{4}{6( p-1)( p-2) n^{p-3}}$

Edit: As correctly pointed out in the comments, this series diverges for $p = 4$ and later on converges for $p = 5$ (the famous Basel problem).

This is because when $p = 4$, each term effectively becomes of the form $\frac 1n$ and this we know to be diverging. On the other hand for $p = 5$ it becomes $\frac 1{n^2}$.

mrtechtroid
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  • For $p=4$ we can bound the terms from below by $\frac 1{3n}$, the sum of which diverges as the harmonic series. – Ross Millikan Jan 15 '23 at 14:34
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    The sequence converging to 0 is necessary but not sufficient for saying a series converges, so I think this answer is wrong – ChristmasTree Jan 15 '23 at 14:50
  • @ChristmasTree Hi there, I realized just now that each term of the sequence was the term shown, and I forgot completely, the part that this term was supposed to be further added. I'll correct my answer. – mrtechtroid Jan 15 '23 at 15:07
  • @mrtechtroid Thanks for your time. I didn't know the case for p=5 was well known, can you give any references on that? – ChristmasTree Jan 15 '23 at 17:27
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    Hi there, here's a question asking the same: https://math.stackexchange.com/questions/1954692/why-does-the-sum-of-inverse-squares-equal-pi2-6..... We generally prove this using Fourier Series – mrtechtroid Jan 15 '23 at 23:36
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    Here's the Wiki Article explaining in depth about the same: https://en.wikipedia.org/wiki/Basel_problem – mrtechtroid Jan 15 '23 at 23:37