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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.

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    $\begingroup$ Well you have $n$ copies of $1/n$ so overall you have $n \cdot 1/n = 1$ $\endgroup$ – Zubzub Mar 9 '17 at 20:53
  • $\begingroup$ @Zubzub Of course -- it seems so obvious now. Thanks for the assistance. $\endgroup$ – The Pointer Mar 9 '17 at 20:54
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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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To augment the other answers, the summation $$\sum_{j=1}^n{1\over n}$$ only depends on $j$, of which there are no $j$s there. So we may pull the $\displaystyle {1\over n}$ out of the sum to give us $$\sum_{j=1}^n{1\over n} = {1\over n} \sum_{j=1}^n(1) = {1\over n}(\underbrace{1+1+\cdots +1}_{n \text{ times}}) = {1\over n}(1\times n) = {1\over n}(n) = 1.$$

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  • $\begingroup$ Notably here we are applying the distributive property in the first step $\endgroup$ – Stella Biderman Mar 9 '17 at 22:42

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