$\displaystyle \sum_{n=2}^x {1\over{x}} (1+{1\over{x}})^{n-2}$ anyone have a way of making this prettier?
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Originally Posted by jbpellerin $\displaystyle \sum_{n=2}^x {1\over{x}} (1+{1\over{x}})^{n-2}$ anyone have a way of making this prettier? $\displaystyle \sum_{n=2}^x {1\over{x}} (1+{1\over{x}})^{n-2}$ is equivalent to $\displaystyle \sum_{n=2}^x \frac {1}{x}(\frac{x+1}{x})^{n-2}$ $\displaystyle \sum_{n=2}^x \frac{(x+1)^{n-2}}{{x}^{n-1}}$ Can you take it from there?
actually it turned out to be pretty easy $\displaystyle (1+{1\over{x}})^{x-1} -1$ it's a geometric series
First observation : you can factor out 1/x from the sum : $\displaystyle =\frac 1x \sum_{n=2}^x \left(1+\frac 1x\right)^{n-2}$ Second observation : what you said > geometric series !
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