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I hava an infinite sum

$$\sum_{n=0}^{\infty}\frac{1}{3n^{2}+4n+1}$$

I factored the denominator

$$\sum_{n=0}^{\infty}\frac{1}{\left(3n+1\right)\left(n+1\right)}$$

Then I separated the fraction

$$\frac{1}{2}\sum_{n=0}^{\infty}\frac{3}{\left(3n+1\right)}-\frac{1}{\left(n+1\right)}$$

Then I set 1 (the numerator) to be equal to x to some power which I don't know if I can do

$$\frac{3}{2}\sum_{n=0}^{\infty}\frac{x^{3n+1}}{3n+1}-\frac{1}{2}\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1}$$

Then I set the integral which would satisfy the previous terms $$\frac{3}{2}\sum_{n=0}^{\infty}\int_{0}^{1}x^{3n}dx$$

and

$$-\frac{1}{2}\sum_{n=0}^{\infty}\int_{0}^{1}x^{n}dx$$

Then I changed the order of summation and integration and I got $$\frac{3}{2}\int_{0}^{1}\frac{1}{1-x^{3}}dx$$ and $$-\frac{1}{2}\int_{0}^{1}\frac{1}{1-x}dx$$

The first integral can be factored to

$$\frac{3}{2}\int_{0}^{1}\frac{1}{\left(1-x\right)\left(1+x+x^{2}\right)}dx$$

Then separated $$\frac{1}{2}\int_{0}^{1}\frac{1}{1-x}+\frac{x+2}{1+x+x^{2}}dx$$

The first one will cancel out with

$$-\frac{1}{2}\int_{0}^{1}\frac{1}{1-x}dx$$

And I'm left with

$$\frac{1}{2}\int_{0}^{1}\frac{x+2}{1+x+x^{2}}dx$$

which is $$\frac{\sqrt{3}\pi}{12}+\frac{\ln\left(3\right)}{4}$$

But the correct answer is

$$\frac{\sqrt{3}\pi}{12}+\frac{3\ln\left(3\right)}{4}$$

So I would like to ask if this approach is invalid or if I'm just missing something.

lcv
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Bruh
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  • The 1st integral looks strange because $1-x^2 = (1-x)(1+x)$ and not $(1-x)(1+x+x^2)$. – emacs drives me nuts May 18 '22 at 17:21
  • Double-check your partial fraction decomp for the integral, I’m away from paper atm but it doesn’t seem to add up – Stephen Donovan May 18 '22 at 17:22
  • @emacsdrivesmenuts The denominator is $1-x^3,$ so he factored using difference of cubes – Stephen Donovan May 18 '22 at 17:24
  • @Stephen Donovan: Ok, that font was sooo tiny I misread the exponent... My bad. – emacs drives me nuts May 18 '22 at 17:25
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    The problem is that the integral from 0 to 1 of 1/(1-x) doesn't converge, so you are cancelling two divergent integrals. This problem basically arises from the fact that 1/(3n+1) doesn't converge as a sum and nor does 1/(n+1). You can't treat them separately. – latbbltes May 18 '22 at 17:28
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    @latbbltes This is true but it can be avoided by not breaking up the sum in an earlier step (but yes the justification here definitely needs refinement) – Stephen Donovan May 18 '22 at 17:31
  • @StephenDonovan I see what you mean. Is there some explanation in terms of not being able to swap integrals and sum in this particular case maybe? – latbbltes May 18 '22 at 17:35
  • Since $\frac1k=\int_0^1x^{k-1}dx$ for $k>0$, the rigorous version of your idea is$$\frac12\sum_{n=0}^{N-1}\left(\frac{3}{3n+1}-\frac{1}{n+1}\right)=\frac12\int_0^1\left(3\frac{1-x^{3N}}{1-x^3}-\frac{1-x^N}{1-x}\right)dx,$$for which $\lim_{N\to\infty}\int_0^1f_N(x)dx=\int_0^1\lim_{N\to\infty}f_N(x)dx$ doesn't work. – J.G. May 18 '22 at 18:08
  • Could I ask then, what would be the legitimate approach, since this whole thing doesn't work? – Bruh May 18 '22 at 18:15

3 Answers3

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$$ \begin{align}\sum_{n=0}^{\infty}\frac{1}{3n^{2}+4n+1} &= \frac{1}{2} \sum_{n=0}^{\infty} \left(\frac{3}{3n+1}-\frac{1}{n+1} \right) \\ &= \frac{1}{2} \sum_{n=0}^{\infty} \left(3\int_{0}^{1}x^{3n} \, \mathrm dx- \color{red}{3\int_{0}^{1} x^{3n+2} \, \mathrm dx}\right) \\ &= \frac{3}{2} \sum_{n=0}^{\infty} \int_{0}^{1} \left(x^{3n}-x^{3n+2} \right) \, \mathrm dx \\ &= \frac{3}{2} \sum_{n=0}^{\infty} \int_{0}^{1} x^{3n}\left(1-x^{2}\right) \, \mathrm dx \end{align}$$

Since $x^{3n}(1-x^{2})$ is nonnegative for all $n$ and all $x$, Tonelli's theorem allows us to interchange the order of integration and summation.

Therefore, we have $$ \begin{align} \sum_{n=0}^{\infty}\frac{1}{3n^{2}+4n+1} &= \frac{3}{2} \int_{0}^{1} (1-x^{2}) \sum_{n=0}^{\infty} x^{3n}\, \mathrm dx \\ &= \frac{3}{2} \int_{0}^{1} \frac{1-x^{2}}{1-x^{3}} \, \mathrm dx \\ &= \frac{3}{2} \int_{0}^{1} \frac{x+1}{x^{2}+x+1} \, \mathrm dx \\ &= \frac{3}{4} \int_{0}^{1} \left( \frac{2x+1}{x^{2}+x+1} +\frac{1}{x^{2}+x+1} \right)\, \mathrm dx \\ &= \frac{3}{4} \, \ln(3) + \frac{3}{4} \int_{0}^{1} \frac{\mathrm dx}{\left(x+\frac{1}{2} \right)^{2} + \frac{3}{4}} \, \mathrm dx \\ &= \frac{3}{4} \, \ln(3) + \frac{3}{2\sqrt{3}} \, \arctan \left(\frac{2}{\sqrt{3}} \left( x+\frac{1}{2}\right) \right) \Bigg|_{0}^{1} \\ & = \frac{3}{4} \, \ln(3) + \frac{3}{2 \sqrt{3}} \left(\frac{\pi}{3}- \frac{\pi}{6} \right) \\ &=\frac{3}{4} \, \ln(3) + \frac{\sqrt{3} \pi}{12}. \end{align}$$

Random Variable
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  • Thanks for an answer. Since you posted something very similar to mine, but yours correct, can I ask why there needs to be x to the power of 3n instead only n? – Bruh May 19 '22 at 18:03
  • @Bruh Because Tonelli's theorem can't be applied to $\frac{1}{2} \sum_{n=0}^{\infty} \int_{0}^{1} \left(3x^{3n}-x^{n} \right) , \mathrm dx$ since $3x^{3n}-x^{n}$ is not nonnegative on the interval $[0,1]$. It changes sign. The reason you got an incorrect result was because switching the order of integration and summation wasn't justified. – Random Variable May 19 '22 at 19:43
  • Right, I understand now. Thanks a lot. – Bruh May 19 '22 at 20:29
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    Nice approach and clear explanation, why we should be careful with substitutions - having the difference of two diverging sums, we can easily get an additional finite piece. I would say, we can change the order of summation and integration, if we guarantee the same number of every partial sum - what you did in fact, choosing the appropriate view of the second integral (marked red). We could also choose $\frac{1}{n+1/3}-\frac{1}{n+1}=\int_0^1(t^{n-2/3}-t^n)dt$ and then make further changes of the variable. It is also important to make the same change of the variable in both integral. – Svyatoslav May 19 '22 at 21:04
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If you're looking to rewrite the series as an integral, here's a way of doing it:

Consider

$$\sum_{n=0}^{\infty}{1\over{(3n+1)(n+1)}}x^{3n+1}$$

So that its derivative will cancel out one of the factors in the denominator:

$$\sum_{n=0}^{\infty}{1\over{(n+1)}}x^{3n}$$

Similar to how you showed, we can now write the original series as a definite integral:

$$\sum_{n=0}^{\infty}{1\over{(3n+1)(n+1)}}=\sum_{n=0}^{\infty}\int_{0}^{1}{1\over{(n+1)}}x^{3n}dx $$ $$ =\int_{0}^{1}\sum_{n=1}^{\infty}{(x^3)^{n-1}\over n} dx=\int_{0}^{1}{1\over x^3}\sum_{n=1}^{\infty}{(x^3)^n\over n} dx$$

The last sum is recognizable as the power series expansion of $-\ln(1-x^3)$, so that:

$$ \sum_{n=1}^{\infty}{1\over {(3n+1)(n+1)}} =-\int_{0}^{1}{1\over x^3}\ln(1-x^3) dx$$

The integral is a little beyond me at the moment, however.

Andrew Sotomayor
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    This integral is a derivative of a Beta function. This gives the final result as $\frac{\psi(1)-\psi(1/3)}{2}$, which is the desired value. – J.G. May 18 '22 at 18:41
  • Thank you very much, I tried to compute this integral and yes, it is the correct result. I used integration by parts which resulted in ln(1-x^3)/(2x^3)+3/2*integral(1/(1-x^3))dx. – Bruh May 18 '22 at 18:47
  • @Bruh right, I'm getting something similar, but the evaluations are a bit tricky. – Andrew Sotomayor May 18 '22 at 19:04
  • I see now, so the first term should not be evaluated, but instead combined with the $\ln(1-x)$ term from the second integral after IBP, all within the limit, which results in $\ln(x^2+x+1)$ and as $x\rightarrow1$ this approaches $\ln 3$ which in turn adds that missing $\ln 3 \over 2$ term from your original answer. – Andrew Sotomayor May 18 '22 at 19:58
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Hi welcome to MSE:As a hint for telescopic series$$\sum_{n=0}^{\infty}\frac{1}{3n^{2}+4n+1} \\=\sum_{n=0}^{\infty}\frac{1}{(3n+1)(n+1)} \\=\sum_{n=0}^{\infty}\frac{3}{(3n+1)(3n+3)} \\=3\sum_{n=0}^{\infty}\frac{(3n+2)}{(3n+1)(3n+2)(3n+3)} \\=3\sum_{n=0}^{\infty}\frac{(3n+3)-1}{(3n+1)(3n+2)(3n+3)} \\=3\sum_{n=0}^{\infty}\frac{1}{(3n+1)(3n+2)}-3\sum_{n=0}^{\infty}\frac{1}{(3n+1)(3n+2)(3n+3)}$$

Khosrotash
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