Evaluation of $\left(\displaystyle \frac{1}{9}+\frac{1}{18}+\frac{1}{30}+\frac{1}{45}+\frac{1}{63}+\cdots\right) $
What I try: I am trying to solve it using Integration,
So we have
$\displaystyle =\int^{1}_{0}\bigg(x^8+x^{17}+x^{29}+x^{44}+x^{62}+\cdots \bigg)\ \mathrm dx$
$\displaystyle =\int^1_0x^{8}\bigg(1+x^9+x^{21}+x^{36}+\cdots \bigg)\ \mathrm dx$
The above series is not in the infinite geometric series.
I did not understand how I could solve it after that.
Please have a look at this; thanks.
Best Answer
I present a separate method not involving calculus.
We try to find a pattern in the sequence $a_n$ with initial terms $9, 18, 30, 45\dots$ and so on. Out of the many patterns one can interpolate, one which is most obvious is that the differences of consecutive terms is in arithmetic progression. More specifically, $a_{n+1}-a_n = 3n+6$. Since the first order of difference of $a_n$ is linear, the general term $a_n$ must be a quadratic.
Thus $a_n = an^2+bn+c$. Then $a+b+c=9$, $4a+2b+c=18$ and $9a+3b+c=30$. Solving we have $a=\frac32$, $b=\frac92$ and $c=3$. Thus we are to figure out $$\sum_{x=1}^{\infty} \frac2{3x^2+9x+6} = \frac23\sum_{x=1}^{\infty} \left(\frac1{x+1} -\frac1{x+2}\right)$$ which telescopes.