What is the Sum of Series: Problem of the Week #153 (March 2nd, 2015)?

[anemone]

For integers $n\ge 1$, determine the sum of $n$ terms of the series

$\dfrac{2n}{2n-1}+\dfrac{2n(2n-2)}{(2n-1)(2n-3)}+\dfrac{2n(2n-2)(2n-4)}{(2n-1)(2n-3)(2n-5)}+\cdots$

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[anemone]

Congratulations to the following members for their correct solutions::)

1. Opalg
2.MarkFL
3. lfdahl

Solution from Opalg:

Spoiler

Proof by induction that \(\displaystyle \frac{2n}{2n-1} + \frac{2n(2n-2)}{(2n-1)(2n-3)} + \frac{2n(2n-2)(2n-4)}{(2n-1)(2n-3)(2n-5)} + \ldots + \frac{2n(2n-2) \cdots 2}{(2n-1)(2n-3) \cdots 1} = 2n.\)

Base case $n=1$: the left side is $\frac21$ and the right side is $2$.

Inductive step: Suppose that the result holds for $n$. Then $$ \frac{2n+2}{2n+1} + \frac{(2n+2)2n}{(2n+1)(2n-1)} + \frac{(2n+2)2n(2n-2)}{(2n+1)(2n-1)(2n-3)} + \ldots + \frac{(2n+2)2n(2n-2) \cdots 2}{(2n+1)(2n-1)(2n-3) \cdots 1}$$ $$\hspace{2em}= \frac{2n+2}{2n+1} \biggl(1 + \frac{2n}{2n-1} + \frac{2n(2n-2)}{(2n-1)(2n-3)} + \ldots + \frac{2n(2n-2) \cdots 2}{(2n-1)(2n-3) \cdots 1} \biggr)$$ $$\hspace{2em}= \frac{2n+2}{2n+1}\bigl(1 + 2n\bigr) = 2n+2.$$ This says that the result is true for $n+1$ and completes the inductive proof.