Saturday, 1 February 2014

summation - Why is $sum_{j = 1}^{n} dfrac{1}{n} = 1$



I encountered $\sum_{j = 1}^{n} \dfrac{1}{n} = 1$ in my textbook, but I do not understand why this summation $= 1$. My textbook provides no reasoning as to why this is the case.




My understanding is that, since there is nothing in $\dfrac{1}{n}$ that depends on $j$, it seems that we are just summing $\dfrac{1}{n}$ to itself up to $n$. However, I'm not sure how to interpret this and how it equals a value of $1$.



I apologise if there is already a question on this, but my searches have encountered nothing that addresses this specific summation. If I am mistaken, I would appreciate it if someone could please redirect me.



I would greatly appreciate it if people could please take the time to explain the reasoning behind this.


Answer



Note that
$$
\sum_{j=1}^n \frac 1n = \overbrace{\frac 1n + \frac 1n + \cdots + \frac 1n}^{n\text{ times}} = n \cdot \frac 1n = 1
$$



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