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Product of Two Kronecker delta

Writer Sophia Terry
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Assume that the range of dummy indices is from 1 to N

$$\delta_{ij} \delta_{jn} = \delta_{i1} \delta_{1n} + \delta_{i2} \delta_{2n} + \delta_{i3} \delta_{3n} +\cdots + \delta_{ii} \delta_{in} + \cdots \delta_{in} \delta_{nn} + \cdots + \delta_{iN} \delta_{Nn} $$

Assuming

$$i = n$$

we get

$$\delta_{ij} \delta_{jn} = \delta_{ii} \delta_{in} + \delta_{in} \delta_{nn}$$

Substituting i = n, we have

$$\delta_{ij} \delta_{jn} = \delta_{ii} \delta_{ii} + \delta_{ii} \delta_{ii} = 2$$

I know this is wrong and $$\delta_{ij} \delta_{jn} = \delta_{in}$$ but I don't know where I am making a mistake.

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