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Question

Transform each by making the change of variable i=k+1.

k=1nkk2+4\prod _ { k = 1 } ^ { n } \frac { k } { k ^ { 2 } + 4 }

Solution

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Given:

k=1nkk2+4\prod_{k=1}^n \dfrac{k}{k^2+4}

i=k+1i=k+1

Let us first rewrite the sum:

k=1nkk2+4=k=1n(k+1)1((k+1)1)2+4k=(k+1)1\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 kk takes on values from 1 to nn, then k+1k+1 needs to take on values from 1+1=21+1=2 to n+1n+1

=k+1=2n+1(k+1)1((k+1)1)2+4\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+1k+1 by ii

=i=2n+1i1(i1)2+4\begin{align*} =\prod_{i=2}^{n+1} \dfrac{i-1}{(i-1)^2+4} \end{align*}

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