Tricky Sum!

Algebra Level 4

$\sum_{n=2}^\infty \dfrac1{(n^2-1)n^2} = \dfrac{1}{3\cdot4}+\dfrac{1}{8\cdot 9}+\dfrac{1}{15\cdot16}+\dfrac{1}{24\cdot25}+\cdots$

The series above is equal to $\dfrac{A-B\pi^{C}}{D}$ where $A,B,C$ and $D$ are positive integers with $A$ and $B$ coprime.

Find the value of $A+B+C+D$ .

The answer is 37.

1 solution

Isaac Buckley
Dec 18, 2015

$S=\sum_{n=2}^{\infty} \frac{1}{(n^2-1)n^2}$

Via partial fractions we can split this into a telescopic series and a well known Dirichlet L-Series

$S=\left[\frac{1}{2}\sum_{n=2}^{\infty}\left(\frac{1}{n-1}-\frac{1}{n+1}\right)\right]-\left[-1+\sum_{n=1}^{\infty} \frac{1}{n^2}\right] =\frac{1}{2}\left(1+\frac{1}{2}\right)+1-\frac{\pi^2}{6} =\frac{21-2\pi^2}{12}$

Therefore $A+B+C+D=21+2+2+12=\boxed{37}$

Did the same way. It may be noted that $\dfrac 1 {n^2}$ can not be split into partial fractions but Dirichlet L-Series is used. The L-Series starts from 1, we from 2. It is because of this $\left[-1+\sum_{n=1}^{\infty} \frac{1}{n^2}\right]$ is used.

Niranjan Khanderia - 5 years, 5 months ago

Dirichlet L-Series by definition start from $n=1$ .

That's why i changed $\sum\limits_{n=2}^{\infty} \frac{1}{n^2}$ to $-1+\sum\limits_{n=1}^{\infty} \frac{1}{n^2}$ .

Isaac Buckley - 5 years, 5 months ago

I totally agree. I had mention it so those not familiar may make a note of it. I have improved my comment.

Niranjan Khanderia - 5 years, 5 months ago

Did the same way!

Shreyash Rai - 5 years, 5 months ago

Tricky Sum!

The answer is 37.

1 solution

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