COMPUTE \displaystyle \sum _{k=1}^{\infty } \sum _{t=1}^{T} 0.5^ k=\quad
To compute the value of the given series, we can first evaluate the inner sum:
\[\sum _{k=1}^{\infty } 0.5^ k\]
This is a geometric series with a common ratio of 0.5. The formula for the sum of an infinite geometric series is \(\frac{a}{1-r}\), where \(a\) is the first term and \(r\) is the common ratio.
In this case, the first term \(a\) is 0.5 and the common ratio \(r\) is 0.5. So the sum of this infinite geometric series is:
\[\frac{0.5}{1-0.5} = \frac{0.5}{0.5} = 1\]
Now, we can substitute this value back into the original series:
\[\sum _{k=1}^{\infty } \sum _{t=1}^{T} 0.5^ k = \sum _{t=1}^{T} 1\]
Since the value of the inner sum does not depend on \(t\), we can simply multiply 1 by \(T\) to get the final result:
\[\sum _{k=1}^{\infty } \sum _{t=1}^{T} 0.5^ k = T\]
To compute the sum \(\displaystyle \sum _{k=1}^{\infty } \sum _{t=1}^{T} 0.5^ k\), we can first evaluate the inner sum \(\displaystyle \sum _{t=1}^{T} 0.5^ k\) for each value of \(k\), and then sum up these values for all values of \(k\) from 1 to infinity.
The inner sum \(\displaystyle \sum _{t=1}^{T} 0.5^ k\) can be simplified as follows:
\[
\sum _{t=1}^{T} 0.5^ k=T\cdot 0.5^ k
\]
Now, we need to compute the sum \(\displaystyle \sum _{k=1}^{\infty } T\cdot 0.5^ k\). This is a geometric series with a common ratio of \(0.5\) and the first term \(T\). The sum of a geometric series can be computed using the formula:
\[
\sum _{k=1}^{\infty } ar^{ k} =\frac{a}{1-r}
\]
Applying this formula, we get:
\[
\sum _{k=1}^{\infty } T\cdot 0.5^ k =\frac{T}{1-0.5} =2T
\]
Therefore, the sum \(\displaystyle \sum _{k=1}^{\infty } \sum _{t=1}^{T} 0.5^ k\) is equal to \(2T\).