# series: $\frac{1}{2\cdot 3\cdot 4}+\frac{1}{4\cdot 5\cdot 6}+\frac{1}{6\cdot 7\cdot 8}+\cdots$ [duplicate]

We have the infinite series:$$\frac{1}{2\cdot 3\cdot 4}+\frac{1}{4\cdot 5\cdot 6}+\frac{1}{6\cdot 7\cdot 8}+\cdots$$

This is not my series: $\frac{1}{1\cdot 2\cdot 3}+\frac{1}{2\cdot 3\cdot 4}+\frac{1}{3\cdot 4\cdot 5}+\frac{1}{4\cdot 5\cdot 6}+\cdots$ so I cannot use $\sum_{k=1}^n \frac{1}{k(k+1)(k+2)}$

My attempt:

I know that this type of series solved by making it telescoping series but here, I am unable find general term of the series. Thank you.

• math.stackexchange.com/questions/2108596/… – lab bhattacharjee Jan 23 '17 at 14:44
• My series is not like this:$\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+\frac{1}{4.5.6}+\cdots$. – MatheMagic Jan 23 '17 at 14:48
• I think to find $\frac{1}{2\cdot3\cdot4}+\frac{1}{5\cdot6\cdot7}+\frac{1}{8\cdot9\cdot10}+...$ is much interesting. – Michael Rozenberg Jan 23 '17 at 14:50

$$\dfrac2{2n(2n+1)(2n+2)}=\dfrac{2n+2-2n}{2n(2n+1)(2n+2)}=\dfrac1{2n(2n+1)}-\dfrac1{(2n+1)(2n+2)}$$

$$=\dfrac1{2n}-\dfrac1{2n+1}-\left(\dfrac1{2n+1}-\dfrac1{2n+2}\right)$$

$$\sum_{n=2}^\infty\left(\dfrac1{2n}-\dfrac1{2n+1}-\left(\dfrac1{2n+1}-\dfrac1{2n+2}\right)\right)$$

$$=\left(\dfrac12-\dfrac13+\dfrac14-\dfrac15+\cdots\right)-\left(\dfrac13-\dfrac14+\dfrac15+\cdots\right)$$

Now $\ln2=1-\dfrac12+\dfrac13-\dfrac14+\cdots$

• It's $3/2-\log4$right? – MatheMagic Jan 23 '17 at 15:08
• @lab bhattacharjee: I think that from line 3 to 4, you changed the order of summation on an infinite number of terms: you sum $\frac 1{2n} - \frac 1 {2n+1}$ on one side, and $\frac{1}{2n+1} - \frac 1 {2n+2}$ on the other. However, this is not legit in the general case (see Riemann series theorem) – Mariuslp Jan 23 '17 at 15:10
• @Mariuslp:then according to you what will be the answer$?$ – MatheMagic Jan 23 '17 at 15:17
• @MatheMagic: My bad, I answered too quickly. This answer is indeed correct. – Mariuslp Jan 24 '17 at 1:53

Let $$f(x)=\sum_{n=1}^\infty\frac1{n(n+1)(n+2)}x^{n+2}.$$ Then $$f'(x)=\sum_{n=1}^\infty\frac1{n(n+1)}x^{n+1},f''(x)=\sum_{n=1}^\infty\frac1{n}x^{n},f'''(x)=\sum_{n=1}^\infty x^{n-1}=\frac1{1-x}.$$ So $$f''(x)=-\ln(1-x)$$ and $$f(1)=-\int_0^1\int_0^x\ln(1-t)dtdx=\cdots$$