The series $\sum\frac{\lfloor n \cdot 2^{n}}{n^{n}}$ is

Question

Accepted Answer

To determine why the series $\sum \frac{\lfloor n \cdot 2^{n}}{n^{n}}$ is convergent, we need to analyze the behavior of the terms $\frac{\lfloor n \cdot 2^{n}}{n^{n}}$ as $n$ approaches infinity.

1. **Explanation for Option 1 (Convergent):**

The term $\frac{\lfloor n \cdot 2^{n}}{n^{n}}$ can be approximated by $\frac{n \cdot 2^{n}}{n^{n}}$ since the floor function $\lfloor x \rfloor$ is very close to $x$ for large $n$. Therefore, we consider the expression $\frac{n \cdot 2^{n}}{n^{n}}$.

Simplifying this, we have:
   $$
   \frac{n \cdot 2^{n}}{n^{n}} = \frac{n}{n^{n-1}} \cdot \left(\frac{2}{n}\right)^{n}
   $$

As $n$ becomes very large, $\left(\frac{2}{n}\right)^{n}$ tends to zero because $\frac{2}{n} < 1$ and raising a number less than 1 to an increasingly large power results in a value approaching zero. Additionally, $\frac{n}{n^{n-1}}$ also tends to zero because $n^{n-1}$ grows much faster than $n$.

Therefore, each term $\frac{n \cdot 2^{n}}{n^{n}}$ approaches zero as $n$ increases, and the series converges.

2. **Explanation for Option 2 (Divergent):**

The series is not divergent because, as shown above, the terms $\frac{n \cdot 2^{n}}{n^{n}}$ approach zero as $n$ increases. For a series to be divergent, the terms must not approach zero or must not satisfy the conditions for convergence. Since the terms do approach zero, the series is not divergent.

3. **Explanation for Option 3 (Conditionally Convergent):**

A series is conditionally convergent if it converges, but the series of absolute values diverges. In this case, the terms $\frac{n \cdot 2^{n}}{n^{n}}$ are positive, and their absolute values are the same as the original terms. Since the series converges absolutely (as shown in the explanation for Option 1), it cannot be conditionally convergent.

4. **Explanation for Option 4 (Neither Convergent nor Divergent):**

A series must either converge or diverge; there is no middle ground. Since we have established that the series converges, it cannot be classified as neither convergent nor divergent.

In conclusion, the series $\sum \frac{\lfloor n \cdot 2^{n}}{n^{n}}$ is convergent, supporting Option 1 as the correct answer.