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Question

# Transform each by making the change of variable i=k+1.$\prod _ { k = 1 } ^ { n } \frac { k } { k ^ { 2 } + 4 }$

Solution

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Given:

$\prod_{k=1}^n \dfrac{k}{k^2+4}$

$i=k+1$

Let us first rewrite the sum:

\begin{align*} \prod_{k=1}^n \dfrac{k}{k^2+4}&=\prod_{k=1}^n \dfrac{(k+1)-1}{((k+1)-1)^2+4} &\color{#4257b2}k=(k+1)-1 \end{align*}

Next, we also note that if $k$ takes on values from 1 to $n$, then $k+1$ needs to take on values from $1+1=2$ to $n+1$

\begin{align*} =\prod_{k+1=2}^{n+1} \dfrac{(k+1)-1}{((k+1)-1)^2+4} \end{align*}

Finally, replace every occurrence of $k+1$ by $i$

\begin{align*} =\prod_{i=2}^{n+1} \dfrac{i-1}{(i-1)^2+4} \end{align*}

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