Limiting sigma

Calculus Level 3

r = 1 r 1 × 3 × 5 × 7 × × ( 2 r + 1 ) = ? \large \sum_{r=1}^\infty \dfrac r{1\times3\times5\times7\times\cdots\times(2r+1)} = \ ?

1 3 \frac {1}{3} 2 3 \frac {2}{3} 2 5 \frac {2}{5} 1 2 \frac {1}{2}

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1 solution

Otto Bretscher
Oct 12, 2015

One can easily show, by induction, that the partial sum up to r r is 1 2 1 2 ( 1 3 5 . . . . ( 2 r + 1 ) ) \frac{1}{2}-\frac{1}{2(1*3*5*....*(2r+1))} . This sum goes to 1 2 \boxed{\frac{1}{2}} as r r\rightarrow\infty .

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