The expression in the image is:

$\lim_{{n \to \infty}} \sum_{{t=1}}^n \frac{1}{4^t}$

To evaluate this, let’s examine the sum. The term inside the summation is a geometric sequence, with each term given by $\frac{1}{4^t}$.

The sequence is:

$\frac{1}{4}, \frac{1}{4^2}, \frac{1}{4^3}, \dots, \frac{1}{4^n}$

This is a geometric series with the first term $a = \frac{1}{4}$ and common ratio $r = \frac{1}{4}$. The sum of the first $n$ terms of a geometric series with $|r| < 1$ is given by:

$S_n = \frac{a(1 - r^n)}{1 - r}$

Substituting $a = \frac{1}{4}$ and $r = \frac{1}{4}$:

$S_n = \frac{\frac{1}{4} \left(1 - \left(\frac{1}{4}\right)^n\right)}{1 - \frac{1}{4}} = \frac{\frac{1}{4} \left(1 - \frac{1}{4^n}\right)}{\frac{3}{4}} = \frac{1 - \frac{1}{4^n}}{3}$

Now, we take the limit as $n \to \infty$:

$\lim_{{n \to \infty}} S_n = \lim_{{n \to \infty}} \frac{1 - \frac{1}{4^n}}{3} = \frac{1 - 0}{3} = \frac{1}{3}$

So,

$\lim_{{n \to \infty}} \sum_{{t=1}}^n \frac{1}{4^t} = \frac{1}{3}$

The answer is $\frac{1}{3}$.

Would you like more details or have any questions?

Here are 5 related questions you might be interested in:

1. How does the common ratio affect the convergence of a geometric series?
2. What is the formula for the sum of an infinite geometric series?
3. How can limits be used to evaluate convergent infinite series?
4. What are the applications of geometric series in real-world problems?
5. How does the convergence of a geometric series change when the common ratio is greater than or equal to 1?

Tip: In geometric series, if the common ratio $r$ satisfies $|r| < 1$, the series converges to a finite sum.