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my motivation is to find a "closed form" (may contain integrals) of the general infinite sum $$\sum_{n=0}^\infty\frac{1}{(2n+2k+1)(4n+2k+3)}$$, where k is an arbitrary natural number. The sum can be rewritten as $$\frac{1}{1-2k}\sum_{n=0}^\infty\frac{1-2k}{(2n+2k+1)(4n+2k+3)}$$ which is in turn equals$$\frac{1}{1-2k}\sum_{n=0}^\infty\frac{(4n+2k+3)-2(2n+2k+1)}{(2n+2k+1)(4n+2k+3)}=$$

$$\frac{1}{1-2k}\sum_{n=0}^\infty(\frac{1}{(2n+2k+1)}-\frac{2}{(4n+2k+3)})$$ The individual terms of the sum can be rewritten using integrals $$\frac{1}{1-2k}\sum_{n=0}^\infty\int_0^1(x^{2n+2k}-x^{4n+2k+2})dx$$ Now one could swap the summation and integration signs to obtain $$\frac{1}{1-2k}\int_0^1\sum_{n=0}^\infty(x^{2n+2k}-2x^{4n+2k+2})dx$$

As a test I have put both the original and this expression into Desmos (with 10000 terms and k=1) and both sums appear to be equal to approx. 0.132, so it seems that in this case such a swap is allowed. The thing is, one could use the Taylor series $$\sum_{n=0}^\infty{x^{2n}}=\frac{1}{1-x^{2}}$$ and $$\sum_{n=0}^\infty{x^{4n}}=\frac{1}{1-x^{4}}$$ to simplify it even further to $$\frac{1}{1-2k}\int_0^1(\frac{x^{2k}}{1-x^2}-\frac{2x^{2k+2}}{1-x^4})dx$$, but for some reason, this integral is equal to -0,215 (according to Desmos), which obviously doesn't make sense, since the original sum must be positive. However, I don't know where I made the mistake in the conversion from Taylor series to $$\frac{1}{1-x^{n}}$$

Jakub Peťka
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  • At the very end, do you know when you can interchange summation and integral signs? In particular, note that there's a pole at $ x = 1$, so can we integrate that? (For "simplicity", let's use the Lebesgue integral that allows us to ignore the singularity at $x=1$.) $\quad$ When you're taking the difference, the pole residues happen to cancel out giving you a numerical value. But if the process is invalid, then you can't trust the resultant number. – Calvin Lin Apr 15 '25 at 15:55
  • I'm not aware of any interchange of summation and integral signs at the end, I only did it once in the middle and it appears to be allowed, considering the same numerical result. The pole at x=1 can be removed though, since (1/1-x^4) can be split into (1/2)(1/1-x^2)+(1/2)(1/1+x^2) and if one subtracts the resulting (x^(2k+2)/1-x^2) from x^2k/(1-x^2), the denominator can be canceled out a part of the numerator, leaving just x^2k. Thus, the function inside of the integral doesn't have any poles and as such they cannot be the reason for this "anomaly". – Jakub Peťka Apr 15 '25 at 16:18
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    I'm asking you to justify $ \lim_{n\rightarrow \infty} \sum_n \int_0^1 (x^{2n+2k} - 2x^{4n+2k+2}) , dx = \int_0^1 \lim_{n\rightarrow \infty} \sum_n ( x^{2n+2k} - 2x^{4n+2k+2}) $. $\quad$ This isn't true in general, and there are certain conditions that need to be satisfied. See this post as an example. – Calvin Lin Apr 15 '25 at 16:33
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    Or, put another way, consider $f_n = x^{2n+2k} - 2x^{4n+2k+2}$. Notice that $f_n(1) = -1$, and so $ f_n < 0 $ on some interval $(x_n, 1]$. Consider $ \sum f_n \rightarrow \frac{x^{2k}}{1-x^2 } - \frac{x^{2k+2 } } { 1 - x^4 } $ at $ x = -1$, what happens there? $\quad$ Do you see how this situation is similar to having an infinite matrix where the row sums are all positive and the column sums are all negative? This is why we need some control over the function - point wise convergence isn't sufficient. This is nuanced, so I commend you for noticing the issue, and trying to figure it out. – Calvin Lin Apr 15 '25 at 16:53

1 Answers1

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Too long for a comment

As @Calvin Lin mentioned in the comment, the answer is that you are not allowed to change the order summation/integration, as soon as you are handling the infinite sum. The receipt is, for example, to consider partial sums $S_N$ and in the end take the limit $N\to\infty$ $$S(k)=\frac{1}{1-2k}\sum_{n=0}^\infty\Big(\frac1{2n+2k+1}-\frac2{4n+2k+3}\Big)$$ $$=\lim_{N\to\infty}\frac{1}{1-2k}\sum_{n=0}^N\Big(\frac1{2n+2k+1}-\frac2{4n+2k+3}\Big)=\lim_{N\to\infty}S_N(k)$$ where we write $S_N$ in the form $$S_N=\frac{1}{1-2k}\sum_{n=0}^N\Big(\frac1{2n+2k+1}-\frac1{2n+k+3/2}\Big)$$ Now we are allowed to present the sum in the form of the integral and change the order of summation/integration $$S_N=\frac{1}{1-2k}\int_0^1dx\sum_{n=0}^Nx^{2n}\left(x^{2k}-x^{k+1/2}\right)$$ $$=\frac1{1-2k}\int_0^1\frac {x^{2k}-x^{k+1/2}}{1-x^2}dx+\frac1{1-2k}\int_0^1\frac {x^{k+5/2+2N}-x^{2k+2N+2}}{1-x^2}dx$$ $$=\frac1{2-4k}\int_0^1\frac{t^{k-1/2}-t^{k/2-1/4}}{1-t}dt+\frac1{2-4k}\int_0^1\frac{t^{k/2+3/4+N}-t^{k+N+1/2}}{1-t}dt$$ $$=\frac1{2-4k}\left(\psi\Big(\frac k2+\frac34\Big)-\psi\Big(k+\frac12\Big)\right)+\frac1{2-4k}\left(\psi\Big(N+k+\frac32\Big)-\psi\Big(N+\frac k2+\frac74\Big)\right)$$ Taking the limit $N\to\infty$; the second term tends to zero, and we are left with $$S(k)=\frac1{2-4k}\left(\psi\Big(\frac k2+\frac34\Big)-\psi\Big(k+\frac12\Big)\right)$$ At $k=1$ $$S(1)=-\frac12\left(\psi\Big(\frac54\Big)-\psi\Big(\frac32\Big)\right)=-\frac12\left(4+\psi\Big(\frac14\Big)-2-\psi\Big(\frac12\Big)\right)=\frac\pi4+\frac{\ln2}2-1=0.131971...$$

Svyatoslav
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  • If you've verified that "The similar question appeared several times on this site", site rules indicate that you should just link to it and close as an abstract duplicate. – Calvin Lin Apr 15 '25 at 17:18
  • @Calvin Lin, unfortunately, I did not manage to find the appropriate link, thugh I saw a couple of similar questions on the site. For the time being, in order to avoid confusion, I removed this phrase from my post. – Svyatoslav Apr 15 '25 at 17:51