Find $\sum_{n=0}^\infty \big( \frac{1}{3n+4}-\frac{3}{3n+2}+\frac{2}{3n+1}\big)$ without power series or integration?

Question

In a calculus, I have to find a direct way to compute the sum $$\sum_{n=0}^\infty \left( \frac{1}{3n+4}-\frac{3}{3n+2}+\frac{2}{3n+1}\right).$$ I tried to reform a partial sum: $$\sum_{n=0}^N a_n = \frac12+3\sum_{n=1}^N \left(\frac{1}{3n+1}-\frac{1}{3n+2}\right).$$ A classic way then is to introduce the missing terms to use Euler's formula. Problem is, I don't know how to deal with the alternating signs.

What I can't do: use power series (the sum comes from one), neither integration (same problem). I need to find a direct way to sum this, that is, a way which only involves classic techniques with series.

Anyone to help me? Thanks.

Answer 1

Update: This uses integration, which is something OP sought to avoid.

$$\begin{split} \sum_{n=0}^\infty \left( \frac{1}{3n+4}-\frac{3}{3n+2}+\frac{2}{3n+1}\right) &=\sum_{n=0}^\infty \int_0^1x^{3n}(x^3-3x+2)dx\\ &= \int_0^1\frac{x^3-3x+2}{1-x^3}dx\\ &= \int_0^1 \left( \frac 3 {x+x^2+1}-1\right)dx\\ &= \left[ 2{\sqrt 3}\arctan\left(\frac{2x+1}{\sqrt 3}\right) - x\right]_0^1 \\ &= 2{\sqrt 3} \left( \frac \pi 3 - \frac \pi 6\right)-1\\ &= \frac{\pi}{\sqrt 3}-1 \end{split}$$

Answer 2

First note that $$\sum_{n=0}^\infty z^n = \frac{1}{1-z}$$ implies $$\sum_{n=1}^\infty \frac{z^n}{n} = \log\frac{1}{1-z} = -\log(1-z),$$ and so \begin{align} \sum_{n=0}^\infty \frac{z^n}{n+1} &= \frac{-\log(1-z)}{z} \tag1\\ \sum_{n=0}^\infty \frac{z^n}{n+2} &= \frac{-\log(1-z)-z}{z^2} \tag2\\ \sum_{n=0}^\infty \frac{z^n}{n+4} &= \frac{-6\log(1-z)-6z-3z^2-2z^3}{6z^4} \tag3 \end{align} Now let $$a_n = \frac{1}{n+4}−\frac{3}{n+2}+\frac{2}{n+1}$$ and apply $(1)$ through $(3)$ to yield $$\sum_{n=0}^\infty a_n z^n = \sum_{n=0}^\infty \frac{z^n}{n+4} − 3\sum_{n=0}^\infty \frac{z^n}{n+2} + 2\sum_{n=0}^\infty \frac{z^n}{n+1} = \frac{(-6+18z^2-12z^3)\log(1-z) - 6z - 3z^2 + 16z^3}{6 z^4}.$$ Finally apply $$\sum_{n=0}^\infty a_{3n} = \sum_{n=0}^\infty a_n\frac{1+\omega^n+\omega^{2n}}{3} = \frac{1}{3}\sum_{n=0}^\infty a_n + \frac{1}{3}\sum_{n=0}^\infty a_n\omega^n + \frac{1}{3}\sum_{n=0}^\infty a_n(\omega^2)^n,$$ where $\omega=\exp(2\pi i/3)$, to obtain $$\frac{1}{3}\cdot\frac{7}{6} + \frac{1}{3}\left(\frac{-25-19 i\sqrt3}{12} + 3 i \sqrt3 \log\left(\frac{3 - i \sqrt3}{2}\right)\right) + \frac{1}{3}\left(\frac{-25 + 19 i\sqrt3}{12} - 3 i \sqrt3 \log\left(\frac{3 + i \sqrt3}{2}\right)\right) = \frac{\pi}{\sqrt3} - 1.$$

Answer 3

(alternate way, using Fourier series)

Let $S=\displaystyle{\sum_{n=1}^\infty\frac{1}{3n+1}-\frac{1}{3n+2}}=\displaystyle{\sum_{n=1}^\infty\frac{u_n}{n}}$, with $u_{3n+1}=1$, $u_{3n+2}=-1$ and $u_{3n}=0$

Let $f$ be the $3$-periodic odd function defined on $[0,3/2]$ by $f(x)=\dfrac\pi3(3/2-x)$. And let's find out its real Fourier series. Since the function is odd, there are only sine terms.

$$b_n=\frac43\int_0^{3/2}\frac\pi3(3/2-x)\sin(2nx\pi/3)\;\mathrm dx=\frac1n$$

Since $f$ is piecewise $C^1$, the Fourier series converges to the regularized $\hat f(x)=\frac12(f(x^+)+f(x^-))$ (Dirichlet's theorem):

$$\hat f(x)=\sum_{n=1}^\infty b_n\sin(2nx\pi/3)=\sum_{n=1}^\infty \frac{\sin(2nx\pi/3)}n$$

At $x=1$, where $f$ is continuous, we get

$$\hat f(1)=f(1)=\frac\pi6=\sum_{n=1}^\infty\frac{\sin(2n\pi/3)}n =\frac{\sqrt3}2\sum_{n=1}^\infty\frac{\frac{2}{\sqrt3}\sin(2n\pi/3)}n=\frac{\sqrt3}2S$$

That is, $S=\dfrac\pi{3\sqrt3}$.

Answer 4

For this specific case, write $$S_p=\sum_{i=1}^3\Bigg[ \sum_{n=0}^p \frac {a_i}{3n+b_i}\Bigg]=\frac 13\sum_{i=1}^3a_i\Bigg[ \psi\left(p+1+\frac{b_i}{3}\right)-\psi \left(\frac{b_i}{3}\right)\Bigg]$$ Using asymptotics $$3S_p=\log(p) \sum_{i=1}^3a_i-\sum_{i=1}^3 a_i \psi \left(\frac{b_i}{3}\right)+\frac 1{6p}\sum_{i=1}^3 a_i(3+2b_i)+O\left(\frac{1}{p^2}\right)$$ So, if the limit exist, we need $\sum_{i=1}^3a_i=0$ and the infinite sum is $$S_\infty=-\frac 13\sum_{i=1}^3 a_i \psi \left(\frac{b_i}{3}\right)$$ Now, if $b_i$ is not a multiple of $3$ $$\psi \left(\frac{b_i}{3}\right)=c_i-\left(\gamma +\frac{3}{2} \log (3)\right)\implies S_\infty=-\frac 13\sum_{i=1}^3 a_i\,c_i$$ and the first $c_i$ are (with $k=\frac{\pi }{2 \sqrt{3}}$) $$\left( \begin{array}{cc} i & c_i \\ 1 & -k \\ 2 & +k \\ 4 & 3-k \\ 5 & \frac{3}{2}+k \\ 7 & \frac{15}{4}-k \\ 8 & \frac{21}{10}+k \end{array} \right)$$

So,what is left is $$ S_\infty=-a_1+\frac{a_1-a_2+a_3}3 k=-a_1-\frac 23 a_2 \,k=-a_1-a_2 \frac{\pi }{3 \sqrt{3}}$$ and then the result.