# Limit of a geometric summation

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.

-
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

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