The given expression involves a sequence with a limit as $n \to \infty$:

$\lim_{n \to \infty} \left( \frac{3}{4} + \frac{5}{16} + \frac{9}{64} + \dots + \left( \frac{1}{4^n} + \frac{2^n}{4^n} \right) \right).$

Let's break this down step by step:

1. The terms appear to follow a pattern:

   • First term: $\frac{3}{4}$
   • Second term: $\frac{5}{16}$
   • Third term: $\frac{9}{64}$
   • General term: $\frac{k^2}{4^k}$, where $k$ is the position of the term (for $k \geq 1$).

2. The limit expression includes an additional term at the end, which is the sum of: $\frac{1}{4^n} + \frac{2^n}{4^n} = \frac{1}{4^n} + \left( \frac{2}{4} \right)^n = \frac{1}{4^n} + \frac{2^n}{4^n}.$ Simplifying the second term: $\frac{2^n}{4^n} = \left( \frac{1}{2} \right)^n.$

3. The entire sequence resembles a series where each term has the form $\frac{k^2}{4^k}$ plus the remaining terms involving powers of $\frac{1}{4}$ and $\frac{1}{2}$.

Limit Calculation:

The series involves a combination of decreasing terms as $n \to \infty$. Let's analyze the behavior:

• The terms $\frac{k^2}{4^k}$ decrease rapidly as $k \to \infty$ because $4^k$ grows much faster than $k^2$, making each successive term smaller.
• The additional terms $\frac{1}{4^n}$ and $\frac{2^n}{4^n}$ tend to 0 as $n \to \infty$, since both powers of $4$ and $\frac{1}{2}$ decay to 0.

Therefore, the entire expression converges to a limit, and since all terms decay to 0, the limit of the sequence is:

$\boxed{1}.$

Would you like more details or have any questions?

Related Questions:

1. How do we prove the convergence of the series $\sum_{k=1}^{\infty} \frac{k^2}{4^k}$?
2. What is the behavior of the general term $\frac{k^2}{4^k}$ as $k \to \infty$?
3. Can the series be summed using any special formulas for power series?
4. How does the inclusion of $\frac{1}{4^n} + \frac{2^n}{4^n}$ affect the convergence?
5. What techniques can be used to find limits of similar sequences involving exponential decay?

Tip:

For sequences involving powers, always check the rate at which terms decay, especially for powers like $4^n$ or $2^n$, as they often lead to zero in limits.