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Can anyone think of a way to simplify

$$ \sum_{k = 1}^n \left[ 1 - \left(\frac{n - 1}{n}\right)^{k - 1} \right] $$

to a more elegant expression? I've been trying to tweak it using the binomial theorem after expanding it but it's just looking uglier and uglier.

Edit:


So far I've been able to rewrite it as

$$ n - \sum_{k = 1}^n \sum_{j = 0}^{k - 1} {k - 1 \choose j}\left(\frac{-1}{n}\right)^j $$

by expanding the summation and making use of the binomial theorem.

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    $\begingroup$ The sum of 1 is easy, and the rest is a geometric progression. $\endgroup$ – Gerry Myerson Nov 10 '13 at 5:27
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With the abbreviation $q=\frac{n-1}{n}$ your sum can be written as $$\sum_{k = 1}^n \left( 1 - \left(\frac{n - 1}{n}\right)^{k - 1} \right) =\sum_{k = 1}^n \left( 1 - q^{k - 1} \right) =\sum_{k = 1}^n 1 - \sum_{k = 1}^n q^{k - 1} = n - \frac{q^n-1}{q-1} $$ The last term is the well-known geometric sum. Now using $$q-1=\frac{n-1}{n}-1= \frac{n-1-n}{n} = -\frac{1}{n}$$ the last expressioncan be simplified as $$ n - \frac{q^n-1}{q-1} = n + n(q^n-1)= nq^n = n \left(\frac{n - 1}{n}\right)^n, $$ and therefore the final result is $$\sum_{k = 1}^n \left( 1 - \left(\frac{n - 1}{n}\right)^{k - 1} \right) = n \left(\frac{n - 1}{n}\right)^n $$

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