# Sum of the serie $\sum_{n=1}^{\infty}\frac{1}{4n^3-n}$

I am trying to calculate the sum of this infinite series $$\sum_{n=1}^{\infty}\frac{1}{4n^3-n}.$$ I only know that $$\frac{1}{4n^3-n}=-\frac{1}{n}+ \frac{1}{2n+1} +\frac{1}{2n-1}.$$ Can you help me, please?

thanks.

• I have tried to use the partial sums of the harmonic series up to $2m$ – Mainkit Apr 10 '17 at 10:36
• Are you supposed to know about generalized harmonic numbers ? – Claude Leibovici Apr 10 '17 at 10:42
• the sum is given by $$-1+2\ln(2)$$ – Dr. Sonnhard Graubner Apr 10 '17 at 10:50
• I don't know about generalized harmonic numbers. – Mainkit Apr 10 '17 at 11:02

$$\dfrac1{4n^3-n}=-\dfrac1{2n}+\dfrac1{2n+1}+\dfrac1{2n-1}-\dfrac1{2n}$$

$$\sum_{n=1}^\infty\dfrac1{4n^3-n}=\sum_{n=1}^\infty\left(-\dfrac1{2n}+\dfrac1{2n+1}+\dfrac1{2n-1}-\dfrac1{2n}\right)$$

$$=-1+2\sum_{r=1}^\infty\dfrac{(-1)^r}r=-1+2\ln(1+1)$$

• I used your help to work with partial sums and get the solution. Thank you very much. – Mainkit Apr 10 '17 at 11:18
• – lab bhattacharjee Apr 10 '17 at 11:29

$$\sum_{n=1}^{N}\frac{1}{4n^3-n}=-\sum_{n=1}^{N}\frac{1}{n}+\sum_{n=1}^{N} \left(\frac{1}{2n+1} +\frac{1}{2n-1}\right)\\=-\sum_{n=1}^{N}\frac{1}{n}+2\sum_{n=0}^{N} \frac{1}{2n+1}-1-\frac{1}{2N+1}\\= -\sum_{n=1}^{N}\frac{1}{n}+2\sum_{n=1}^{2N+1}\frac{1}{n}-2\sum_{n=1}^{N}\frac{1}{2n}-1-\frac{1}{2N+1}$$

Now use $\sum_{n=1}^{N}\frac{1}{n}=\ln(N)+\gamma+o(1)$.

$$\sum_{n=1}^{N}\frac{1}{4n^3-n}=-2\ln(N)+2\ln(2N+1)-1-\frac{1}{2N+1}+o(1)\\=2\ln\left(\frac{2N+1}{N}\right)-1-\frac{1}{2N+1}+o(1)\xrightarrow[N\rightarrow\infty]{}2\ln(2)-1$$

Let $$f(x)=\sum_{n=1}^{\infty}\frac{1}{4n^3-n}x^{2n+1}=\sum_{n=1}^{\infty}\frac{1}{n(2n-1)(2n+1)}x^{2n+1}.$$ Then $$f'(x)=\sum_{n=1}^{\infty}\frac{1}{n(2n-1)}x^{2n}, f''(x)=2\sum_{n=1}^{\infty}\frac{1}{2n-1}x^{2n-1}=\arctan x$$ So $$f(1)=\int_0^1\int_0^x\text{arctanh}t\; dtdx=\int_0^1\int_t^1\arctan tdxdt=\int_0^1(1-t)\text{arctanh} t\; dt=2\ln2-1.$$

Since you do not yet know about generalized harmonic numbers, this is jus added for your curiosity.

These numbers are useful to define partial sums such as $$S_p(a,b)=\sum_{n=1}^p\frac 1{an+b}=\frac 1a ({H_{p+\frac{b}{a}}-H_{\frac{b}a}})$$ and, for large values of $p$, their asymptotic expansion is $$S_p(a,b)=\frac{-H_{\frac{b}{a}}+\log \left({p}\right)+\gamma }{a}+\frac{a+2 b}{2 a^2 p}-\frac{a^2+6 a b+6 b^2}{12 a^3 p^2}+O\left(\frac{1}{p^3}\right)$$ So, for your case, after a shift of index as Jennifer did in the second line of her answer, we have $$T_p=S_p(2,1)+S_p(2,-1)-S(1,0)=H_{p-\frac{1}{2}}-H_p+\frac{1}{2 p+1}-1+\log (4)$$ and the expansion leads to $$T_p=\log (4)-1-\frac{1}{8 p^2}+O\left(\frac{1}{p^3}\right)$$ which shows the limit and how it is approached.

For example $$T_{10}=\frac{44831407}{116396280}\approx 0.385162$$ while the above expansion gives $$T_{10}\approx \log (4)-1-\frac{1}{800}\approx 0.385044$$