Sequences And Series #3

Algebra Level 4

∑ n = 1 ∞ 2 n − 1 3 ⋅ 7 ⋅ 11 ⋯ ( 4 n − 1 ) \large \sum_{n=1}^\infty \frac{2n-1}{3\cdot 7 \cdot 11 \cdots (4n-1)}

Find the value of the expression above.

Inspiration

1 2 \dfrac{1}{2} 1 3 \dfrac{1}{3} 1 1 1 6 \dfrac{1}{6}

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1 solution

The given series can be rewritten in the following way:

1 3 + 1 2 ( 1 3 − 1 3 ∗ 7 ) + 1 2 ( 1 3 ∗ 7 − 1 3 ∗ 7 ∗ 11 ) + 1 2 ( 1 3 ∗ 7 ∗ 11 − 1 3 ∗ 7 ∗ 11 ∗ 15 ) . . . \dfrac{1}{3} +\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{3*7}\right)+\dfrac{1}{2}\left(\dfrac{1}{3*7}-\dfrac{1}{3*7*11}\right)+\dfrac{1}{2}\left(\dfrac{1}{3*7*11}-\dfrac{1}{3*7*11*15}\right) ...

Noticed something ?

Yes , it is a telescoping series .

And we are left with the following two terms :

1 3 + 1 6 = 1 2 \dfrac{1}{3}+\dfrac{1}{6}=\dfrac{1}{2}

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