Evaluate $360 (\frac{1}{15}+\frac{1}{105}+\frac{1}{315}+⋯).$

Question

Evaluate $360$ ($\frac{1}{15}$+$\frac{1}{105}$+$\frac{1}{315}$+$\cdots$).

My Work :-

Well we can clearly see that $15 = 1*3*5$ ; $105 = 3*5*7$ ; $315 = 5*7*9$. So I basically know the pattern, but I want to know how to apply that like maybe we can break $\frac{1}{15}$ into $x*\frac{1}{1}\pm y*\frac{1}{3}\pm z*\frac{1}{5} $ or something like that so that we can cancel some terms in the whole sum.

Answer 1

$T_n = \dfrac{1}{(2n-1)(2n+1)(2n+3)}$

Let $V_n= \dfrac{1}{(2n-1)(2n+1)}=(2n+3)T_n$

$V_{n} - V_{n+1}=\dfrac{1}{(2n-1)(2n+1)}-\dfrac{1}{(2n+1)(2n+3)}=\dfrac{2}{(2n-1)(2n+1)(2n+3)}=2T_n$

Let $\sum_{n=1}^{\infty}T_n=S$

Therefore,

$\sum_{n=1}^{\infty}(V_n - V_{n+1})=(V_1-V_{\infty})=2S$

$V_1=\frac{1}{3},V_{\infty}=0$

Therefore, $S=\frac{1}{6}$ and $360S = 60$

A telescopic series!

Answer 2

Hint: $$\frac{1}{105}=\frac{1}{4}(\frac{1}{3\cdot 5}-\frac{1}{5\cdot 7})$$ $$\frac{1}{315}=\frac{1}{4}(\frac{1}{5\cdot 7}-\frac{1}{ 7\cdot 9})$$ do you observe a pattern........