Sum the infinite series:

Question

$$\frac{5}{3.6}\frac{1}{4^2}+\frac{5.8}{3.6.9}\frac{1}{4^3}+\frac{5.8.11}{3.6.9.12}\frac{1}{4^4}+... \infty $$

$$\frac{5}{3.6}\frac{1}{4^2}+\frac{5.8}{3.6.9}\frac{1}{4^3}+\frac{5.8.11}{3.6.9.12}\frac{1}{4^4}+... < \frac{1}{4^2}+\frac{1}{4^3}+\frac{1}{4^4}+...$$ By comparison test, we can deduce that series is convergent. How to find the exact value?

I tried to make the partial fraction of every coefficient, but It didn't help me to reduce to a known series. Please help me.

See https://math.stackexchange.com/questions/746388/calculating-1-frac13-frac1-cdot33-cdot6-frac1-cdot3-cdot53-cdot6-cdot — lab bhattacharjee, Nov 28 '17 at 05:12
Welcome! Please let me know if u r stuck in understanding the two methods there. — lab bhattacharjee, Nov 28 '17 at 05:19
So it will be $(1-x)^{-5/3}$ at $x = \frac{1}{4}$. We do need to subtract some terms. — jonsno, Nov 28 '17 at 05:25
@labbhattacharjee: What do you think about my method to sum this series :) — Bumblebee, Nov 28 '17 at 05:58
@Bumblebee, https://en.wikipedia.org/wiki/Telescoping_series is a nice tool — lab bhattacharjee, Nov 28 '17 at 06:08
As the denominator of the general term $T_n$ is $\displaystyle3\cdot6\cdot9\cdot12\cdots(3n)=3^n \cdot n!$ . I am not able to factor the numerator. as given in your answer. @labbhattacharjee — , Nov 28 '17 at 06:13
I tried to do the problem using the method suggested by samjoe. there denomenator is not matching. — , Nov 28 '17 at 06:16
@ManeeshNarayanan, Observe that the general term is $$\dfrac12\cdot\dfrac{2\cdot5\cdot8\cdots(3n-1)}{(12)^nn!}$$ — lab bhattacharjee, Nov 28 '17 at 06:19

score 2 · Answer 1 · answered Nov 28 '17 at 05:54

Let $$u_r=\dfrac{5.8.11.\cdots(3r+2)}{3.6.9.12.\cdots (3r+3)}\dfrac{1}{4^{r+1}}=\dfrac{u_{r-1}}{4}\left(\dfrac{3r+2}{3r+3}\right).$$ Now lets find a sequence $(f)$ such that $u_r=f(r-1)-f(r)$ and $f(r)=\dfrac{f(r-1)}{4}\left(\dfrac{3r+2}{3r+3}\right).$ Then we have $$u_r=f(r-1)-\dfrac{f(r-1)}{4}\left(\dfrac{3r+2}{3r+3}\right)=\left(\dfrac{9r+10}{12r+12}\right)f(r-1)$$ and this determines $f$ uniquely. Observe that $$\sum_{r=1}^nu_r=f(0)-f(n)=u_1\left(\dfrac{24}{19}\right)-u_{n+1}\left(\dfrac{12n+24}{9n+19}\right).$$

score 1 · Accepted Answer · answered Nov 28 '17 at 15:03

You may exploit Euler's Beta function. $$\frac{1}{2} \sum_{n\geq 2}\frac{\prod_{k=1}^{n}(3k-1)}{3^n n! 4^n}=\frac{1}{2}\sum_{n\geq 2}\frac{\Gamma\left(n+\tfrac{2}{3}\right)}{\Gamma\left(\tfrac{2}{3}\right)\Gamma(n+1)\,4^n}=\frac{\sqrt{3}}{4\pi}\sum_{n\geq 2}\frac{B\left(\tfrac{1}{3},n+\tfrac{2}{3}\right)}{4^n}\tag{A}$$ and $$\begin{eqnarray*}\frac{\sqrt{3}}{4\pi}\sum_{n\geq 2}\frac{B\left(\tfrac{1}{3},n+\tfrac{2}{3}\right)}{4^n}&=&\frac{\sqrt{3}}{4\pi}\int_{0}^{1}\sum_{n\geq 2}\frac{(1-x)^{-2/3}x^{n-1/3}}{4^n}\,dx\\&=&\frac{\sqrt{3}}{16\pi}\int_{0}^{1}\frac{x^{5/3}}{(1-x)^{2/3}(4-x)}\,dx\end{eqnarray*}\tag{B}$$ can be computed as $\color{red}{\large\frac{4\sqrt[3]{6}-7}{12}}$.

Sorry to annoy you about this quit old question. Somehow today I came across it and notice that my answer of finite sum doesn't give the value in your answer at the limit. However I wasn't able to find a mistake in my answer. Do you see what is happening there? — Bumblebee, Mar 01 '21 at 22:53

Yiorgos S. Smyrlis · Answer 3 · 2017-11-29T08:36:32.853

Hint. Taylor expansion of $f(x)=\sqrt[3]{1+x}$: $$ \left(1-\frac{1}{4}\right)^{-2/3}=1+\frac{2}{3}\cdot\frac{4^{-1}}{1!}+\frac{2}{3}\cdot \frac{5}{3}\frac{4^{-2}}{2!} +\frac{2}{3}\cdot \frac{5}{3}\cdot\frac{8}{3}\frac{4^{-3}}{3!}+\cdots \\ = 1+\frac{1}{6}+2\left(\frac{5}{3\cdot6}\frac{1}{4^2}+\frac{5\cdot8}{3.6.9}\frac{1}{4^3}+\frac{5\cdot8\cdot11}{3\cdot6\cdot9\cdot12}\frac{1}{4^4}+\cdots\right) $$ Thus $$ \frac{5}{3\cdot6}\frac{1}{4^2}+\frac{5\cdot8}{3.6.9}\frac{1}{4^3}+\frac{5\cdot8\cdot11}{3\cdot6\cdot9\cdot12}\frac{1}{4^4}+\cdots=\frac{1}{2}\left(\left(\frac{4}{3}\right)^{2/3}-\frac{7}{6}\right)=\frac{\sqrt[3]{6}}{3}-\frac{7}{12} $$

Sum the infinite series:

3 Answers3