<s>[INST] Evaluate $360 (\frac{1}{15}+\frac{1}{105}+\frac{1}{315}+⋯).$ 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.
 [/INST] $T_n = \dfrac{1}{(2n-1)(2n+1)(2n+3)}$
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!
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