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I am trying to solve the following:

$$ \lim_{n \to \infty} \sum_{k=1}^{nt} \frac{1}{n} \left( 1 - \frac{1}{n} \right)^{k-1}. $$

I think I was able to get the summation correct (see below):

$$ \frac{1}{n} \cdot \frac{1 - \left( 1 - \frac{1}{n} \right)^{nt+1}}{1 - \left(1 - \frac{1}{n}\right)}. $$

However, I am unsure of how to take the limit to infinity of the above summation. I know the answer is supposed to be $1 - e^{-t}$ but I am unsure how to arrive at that.

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What is "nt" in the summation? – user38268 Feb 8 '12 at 4:01
Your second displayed expression simplifies to $1-(1-{t\over nt})^{nt+1}$. I think the power of the $ 1-{1\over n} $ upstairs should be $nt$, not $nt+1$, though. Note $\lim_{k\rightarrow\infty}(1+{\theta\over k})^k=e^\theta$. – David Mitra Feb 8 '12 at 4:06
Just a term we are summing up to. – icobes Feb 8 '12 at 4:06
Thanks David. Makes perfect sense now – icobes Feb 8 '12 at 4:13
up vote 2 down vote accepted

Ignoring the technicality that $nt$ may not be an integer, the sum is, for $t>0$ $$ \sum_{k=1}^{nt} {\textstyle{1\over n}}\bigl(1-{\textstyle{1\over n}}\bigr)^{k-1}= {1\over n}{1-(1-{1\over n})^{(nt+1)-1}\over 1-(1-{1\over n})} =1-(1-{\textstyle{1\over n}})^{nt}. $$

So, $$\eqalign{ \lim_{n\rightarrow\infty } \sum_{k=1}^{nt} {\textstyle{1\over n}}\bigl(1-{\textstyle{1\over n}}\bigr)^{k-1}&= \lim_{n\rightarrow\infty }\bigl[\,1-(1-\textstyle{1\over n})^{nt}\,\bigr]\cr &=\lim_{n\rightarrow\infty }\bigl[\,1-(1-\textstyle{t\over t n})^{nt}\,\bigr]\cr &=\lim_{k\rightarrow\infty }\bigl[\,1-(1-\textstyle{t\over k})^{k}\,\bigr]\cr &=1-e^{-t}. } $$

We really shouldn't ignore the fact that $nt$ may not be an integer, the sum should actually be expressed by $1-(1-{\textstyle{1\over n}})^{\lbrack nt\rbrack}$, say, where $[nt]$ is the integer part of $nt$.

Using the fact that
$$\lim\limits_{k\rightarrow\infty}(1+{\textstyle{t\over k}})^{k+\alpha}= \lim\limits_{k\rightarrow\infty}\bigl[\,(1+{\textstyle{t\over k}})^{k }(1+{\textstyle{t\over k}})^{\alpha }\,\bigr]=e^t, $$ and the squeeze theorem, one can show $$ \lim_{n\rightarrow\infty }\bigl (1-\textstyle{1\over n}\bigr)^{[nt]} =e^{-t}. $$

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