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We want to evaluate the series: $$\sum_{k=1}^{\infty}\frac{9k-4}{3k(3k-1)(3k-2)}$$

My try :

We have : $$\frac{9k-4}{3k(3k-1)(3k-2)}=\frac{1}{3k-1}+\frac{1}{3k-2}-\frac{2}{3k}$$ Therefore:

$$\sum_{k=1}^{\infty}\frac{9k-4}{3k(3k-1)(3k-2)}=\sum_{k=1}^{\infty}\frac{1}{3k-1}+\frac{1}{3k-2}-\frac{2}{3k}=\sum_{k=1}^{\infty}\frac{1}{3k-1}+\frac{1}{3k-2}+\frac{1}{3k}-\sum_{k=1}^{\infty}\frac{1}{k}=\\\sum_{k=1}^{\infty}\frac{1}{k}-\sum_{k=1}^{\infty}\frac{1}{k}=??$$

I need help calculating the sum of this series, thank you for your interest.

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  • $\begingroup$ If you write the numerator as 9𝑘−4=3𝑘+2(3𝑘−2) then the sum becomes two sums similar to ∑1𝑘(𝑘+1). Sums like these have been considered before, see math.stackexchange.com/questions/112161/… $\endgroup$
    – Steen82
    Commented May 12 at 13:15
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    $\begingroup$ The sum, by design, is the specialization of the digamma function's multiplication formula $$\psi(ns)-\frac{1}{n}\sum_{l=0}^{n-1}\psi(s+\tfrac{l}{n})=\ln n$$ to $n=3$, $s=\tfrac{1}{n}$. Any elementary proof for the OP's sum is going to reproduce one of the multitude of elementary proofs for this formula. There a number of homologous questions on this site. $\endgroup$
    – K B Dave
    Commented May 12 at 13:21

2 Answers 2

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Without changing anything to your work, compute the patial summation up to $n$ and thake the limit for $n\to \infty$

$$\sum_{k=1}^{n}\Big(\frac{1}{3k-1}+\frac{1}{3k-2}+\frac{1}{3k}\Big)-\sum_{k=1}^{n}\frac{1}{k}=H_{3 n}-H_n$$

Now, take the asymptotics to get $$ \log (3)-\frac{1}{3 n}+O\left(\frac{1}{n^2}\right)\quad \to \quad \log (3)$$

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By changing indices, $$\sum_{k=1}^\infty\frac{1}{3k-1}+\frac{1}{3k-2}-\frac{2}{3k}=\sum_{k=0}^\infty\frac{1}{3k+2}+\frac{1}{3k+1}-\frac{2}{3k+3}$$

We may split this series into two: $$\sum_{k=0}^\infty\frac{1}{3k+2}+\frac{1}{3k+1}-\frac{2}{3k+3}=\sum_{k=0}^\infty\left(\frac{1}{3k+2}-\frac{1}{3k+3}\right)+\sum_{k=0}^\infty\left(\frac{1}{3k+1}-\frac{1}{3k+3}\right)$$

Notice that $$\sum_{k=0}^n\left(\frac{1}{3k+2}-\frac{1}{3k+3}\right)=\sum_{k=0}^n\int_0^1 x^{3k+1}-x^{3k+2}\,dx=\int_0^1 \sum_{k=0}^n \left(x^{3k+1}-x^{3k+2}\right)\,dx=$$$$\int_{0}^1 (x-x^2)\dfrac{(1-x^{3k+3})}{1-x^3}\,dx$$

By the dominated convergence theorem, we may take the limit and assure that $$\sum_{k=0}^\infty\left(\frac{1}{3k+2}-\frac{1}{3k+3}\right)=\int_{0}^1 \dfrac{x-x^2}{1-x^3}\,dx=\int_{0}^1 \dfrac{x}{1+x+x^2}\,dx$$

This last integral can be solved with standard methods and is equal to $\dfrac{1}{18}(9\ln(3)-\sqrt{3}\pi)$.

A similar approach works for the second sum: $$\sum_{k=0}^n\left(\frac{1}{3k+1}-\frac{1}{3k+3}\right)=\sum_{k=0}^n\int_0^1 x^{3k}-x^{3k+2}\,dx=\int_0^1 \sum_{k=0}^n \left(x^{3k}-x^{3k+2}\right)\,dx=$$$$\int_{0}^1 (1-x^2)\dfrac{(1-x^{3k+3})}{1-x^3}\,dx$$

Again, using the dominated convergence theorem, $$\sum_{k=0}^\infty\left(\frac{1}{3k+1}-\frac{1}{3k+3}\right)=\int_{0}^1 \dfrac{1-x^2}{1-x^3}\,dx=\int_{0}^1 \dfrac{1+x}{1+x+x^2}\,dx=\dfrac{1}{18}(\sqrt{3}\pi+9\ln(3))$$

Putting this two results together, we conclude that $$\boxed{\sum_{k=1}^{\infty}\frac{9k-4}{3k(3k-1)(3k-2)}=\sum_{k=0}^\infty\left(\frac{1}{3k+2}-\frac{1}{3k+3}\right)+\sum_{k=0}^\infty\left(\frac{1}{3k+1}-\frac{1}{3k+3}\right)=\ln(3)}$$

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    $\begingroup$ +1 Nice solution $\endgroup$
    – User
    Commented May 12 at 13:57

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