Evaluate $\sum _{n=1}^{k}\frac{{\left(-4\right)}^{n+2}}{(n-2{)}^{2}}\prod _{i=n}^{2k}\frac{i}{i+1}$ for $k=3$ $\sum _{n=1}^{k}\frac{{\left(-4\right)}^{n+2}}{(n-2{)}^{2}}\prod _{i=n}^{2k}\frac{i}{i+1}$
How can I think this for k=3 ?
Here is my idea: $\prod _{i=n}^{6}\frac{i}{i+1}=\frac{1}{2}.\frac{3}{4}.\frac{4}{5}.\frac{5}{6}.\frac{6}{7}$
then
$\frac{3}{14}\sum _{n=1}^{3}\frac{{\left(-4\right)}^{n+2}}{{\left(n-2\right)}^{2}}=\frac{3}{14}.\left(\frac{{\left(-4\right)}^{3}}{{\left(1-2\right)}^{2}}+\frac{{\left(-4\right)}^{5}}{{\left(3-2\right)}^{2}}\right)=-233,1428571$
Is my idea correct?
 A: *Note: * For $k=1$ to $3$ we have to evaulate the sum for $k\in\{1,2,3\}$:
\begin{align*}
\sum _{n=1}^{k}\frac{{\left(-4\right)}^{n+2}}{(n-2{)}^{2}}\prod _{i=n}^{2k}\frac{i}{i+1}\tag{1}
\end{align*}
Hint: Evaluation at $n=2$ is not feasible, since in that case we have division by zero, which is not admissible. In order to overcome this problem we can exclude the value which causes the indeterminate expressions;.

We consider the  valid expression:
\begin{align*}
\sum _{{n=1}\atop{n\ne 2}}^{k}\frac{{\left(-4\right)}^{n+2}}{(n-2{)}^{2}}
\prod _{i=n}^{2k}\frac{i}{i+1}\tag{2}
\end{align*}
which  can be simplified, since
\begin{align*}
\prod _{i=n}^{2k}\frac{i}{i+1}&=\frac{n}{n+1}\cdot\frac{n+1}{n+2}\cdots\frac{2k}{2k+1 }\\
&=\frac{n}{2k+2}
\end{align*}

A: That's a telescoping product so
$\begin{array}\\
\sum _{n=1}^{k}\dfrac{{\left(-4\right)}^{n+2}}{(n-2{)}^{2}}\prod _{i=n}^{2k}\dfrac{i}{i+1}
&=\sum _{n=1}^{k}\dfrac{{\left(-4\right)}^{n+2}}{(n-2{)}^{2}}\dfrac{n}{2k+1}\\
\end{array}
$
but then there is a problem at $n=2$
when there is a division by zero.
I'll change that to $n+2$
and we get
$\begin{array}\\
\sum _{n=1}^{k}\dfrac{{\left(-4\right)}^{n+2}}{(n+2{)}^{2}}\prod _{i=n}^{2k}\dfrac{i}{i+1}
&=\sum _{n=1}^{k}\dfrac{{\left(-4\right)}^{n+2}}{(n+2{)}^{2}}\dfrac{n}{2k+1}\\
&=\dfrac1{2k+1}\sum _{n=3}^{k+2}\dfrac{{\left(-4\right)}^{n}(n-2)}{n^2}\\
&=\dfrac1{2k+1}\sum _{n=3}^{k+2}\dfrac{{\left(-4\right)}^{n}n}{n^2}-\dfrac1{2k+1}\sum _{n=3}^{k+2}\dfrac{{\left(-4\right)}^{n}2}{n^2}\\
&=\dfrac1{2k+1}\sum _{n=3}^{k+2}\dfrac{{\left(-4\right)}^{n}}{n}-\dfrac{2}{2k+1}\sum _{n=3}^{k+2}\dfrac{{\left(-4\right)}^{n}}{n^2}\\
\end{array}
$
and these can be handeled
by the usual methods.
