I have problems with the series that appear evaluating this integral $$\int_0^n \{x^2\}\,\text{d}x$$ where $\{\cdot\}$ is the fractional part function. Now, I'm pretty sure that $$I=\sum_{k=0}^{n^2-1}\int_\sqrt k^\sqrt{k+1} (x^2-k)\,\text{d}x$$ since $\{x^2\}=x^2-\lfloor x^2\rfloor=x^2-k$ for $x\in[\sqrt k,\sqrt{k+1})$. Now, if that is correct, I proceed integrating and I obtain the series $$I=\frac13\sum_{k=0}^{n^2-1} ((k+1)^\frac32-k^\frac32)-\sum_{k=0}^{n^2-1} k(\sqrt{k+1}-\sqrt k)$$ The first telescopes to $\frac{n^3}3$ while for the second I thought to apply the formula of summation by parts, that is $\sum_{n=k}^Na_nb_n=S_Nb_N-S_{k-1}b_k-\sum_{n=k}^{N-1}S_n(b_{n+1}-b_n)$ where $S_N=\sum_{n=1}^Na_n$. Choosing $a_n=\sqrt{k+1}-\sqrt k$, since its $(n^2-1)$-th partial sum telescopes to $n-1$, the formula gives $$\sum_{k=0}^{n^2-1} k(\sqrt{k+1}-\sqrt k)=(n-1)(n^2-1)-\sum_{k=0}^{n^2-2} (k-1)$$ that seems to be false because $\sum_{k=0}^{n^2-2} (k-1)>(n-1)(n^2-1)$ while $\sum_{k=0}^{n^2-1} k(\sqrt{k+1}-\sqrt k)>0$. I sincerely think I have done something wrong but I don't understand what… Could someone explain or find another solution (the answer given by the book is $-\frac{2n^3}3+\sum_{k=1}^{n^2} \sqrt k$)?

  • $\begingroup$ Do you intend to define $I$ to be the integral in your first display? $\endgroup$ – Eric Towers Apr 11 at 18:15
  • $\begingroup$ Exactly, that is $\endgroup$ – bianco Apr 11 at 18:18
  • $\begingroup$ Is $n$ a nonnegative integer? $\endgroup$ – Eric Towers Apr 11 at 18:30
  • $\begingroup$ Yes, $n$ is defined as a natural number $\endgroup$ – bianco Apr 11 at 19:30

Ok, I just understood my mistake and it's very silly. We have that $S_k=\sqrt{k+1}-1$ and instead I used $S_{k^2-1}$ as $S_k$. Then the answer comes easily expanding, re-elaborating and re-indexing the sums. So stupid...

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