B3. Let $a$ be a real number greater than 1. Sum the infinite series

$$
a^{\ln \frac{1}{a^{0}}}+a^{\ln \frac{1}{a^{1}}}+a^{\ln \frac{1}{a^{2}}}+\ldots
$$

## 56th Mathematical Competition for High School Students in Slovenia  Regional Competition, March 28, 2012

## Solutions for 1st Year

In section A, a correct answer will be awarded two points, while a wrong answer will result in the deduction of one point. To avoid a potential negative final score, each contestant is given an initial 3 points.

| A1 | A2 | A3 |
| :---: | :---: | :---: |
| E | C | B |

## Justifications:
B3. Choose

$$
a^{\ln \frac{1}{a}}=\left(e^{\ln a}\right)^{-\ln a}=\left(\frac{1}{e}\right)^{(\ln a)^{2}}
$$

Since $a \neq 1$, $\ln a \neq 0$, thus $(\ln a)^{2}>0$. Therefore, $\left(\frac{1}{e}\right)^{(\ln a)^{2}}<1$, because $\frac{1}{e}<1$, so $a^{\ln \frac{1}{a}}<1$. It holds that $a^{\ln \frac{1}{a^{n}}}=a^{n \ln \frac{1}{a}}=\left(a^{\ln \frac{1}{a}}\right)^{n}$. Therefore, the given sum is the sum of a geometric series with the first term 1 and the common ratio $a^{\ln \frac{1}{a}}<1$. Thus, this sum is equal to $\frac{1}{1-a^{\ln \frac{1}{a}}}$.

![](https://cdn.mathpix.com/cropped/2024_06_07_66bdf7a61d9bfcb32ca4g-18.jpg?height=72&width=1579&top_left_y=461&top_left_x=273)

![](https://cdn.mathpix.com/cropped/2024_06_07_66bdf7a61d9bfcb32ca4g-18.jpg?height=63&width=1579&top_left_y=525&top_left_x=273)

![](https://cdn.mathpix.com/cropped/2024_06_07_66bdf7a61d9bfcb32ca4g-18.jpg?height=60&width=1579&top_left_y=575&top_left_x=273)

![](https://cdn.mathpix.com/cropped/2024_06_07_66bdf7a61d9bfcb32ca4g-18.jpg?height=54&width=1579&top_left_y=630&top_left_x=273)

Calculation of the sum ........................................................................................................
