$\frac{1}{\lfloor1}+\frac{4}{\lfloor2}+\frac{7}{\lfloor3}+\frac{10}{\lfloor4}+.....\infty=$

Question

Accepted Answer

To solve the given series and justify why Option 2 ($e+2$) is the correct answer, we need to analyze the structure of the series and its convergence.

The series given is:

$$
\frac{1}{\lfloor1} + \frac{4}{\lfloor2} + \frac{7}{\lfloor3} + \frac{10}{\lfloor4} + \ldots
$$

This series can be rewritten in a more general form as:

$$
\sum_{n=1}^{\infty} \frac{3n - 2}{n!}
$$

This is because the numerator of each term follows the arithmetic sequence $1, 4, 7, 10, \ldots$, which can be expressed as $3n - 2$ for $n = 1, 2, 3, \ldots$.

Now, let's break down the series:

$$
\sum_{n=1}^{\infty} \frac{3n - 2}{n!} = \sum_{n=1}^{\infty} \frac{3n}{n!} - \sum_{n=1}^{\infty} \frac{2}{n!}
$$

The first part of the series is:

$$
\sum_{n=1}^{\infty} \frac{3n}{n!} = 3 \sum_{n=1}^{\infty} \frac{n}{n!}
$$

We know that:

$$
\sum_{n=1}^{\infty} \frac{n}{n!} = e
$$

Therefore:

$$
3 \sum_{n=1}^{\infty} \frac{n}{n!} = 3e
$$

The second part of the series is:

$$
\sum_{n=1}^{\infty} \frac{2}{n!} = 2 \sum_{n=1}^{\infty} \frac{1}{n!}
$$

We know that:

$$
\sum_{n=0}^{\infty} \frac{1}{n!} = e
$$

Thus:

$$
\sum_{n=1}^{\infty} \frac{1}{n!} = e - 1
$$

Therefore:

$$
2 \sum_{n=1}^{\infty} \frac{1}{n!} = 2(e - 1) = 2e - 2
$$

Putting it all together, we have:

$$
\sum_{n=1}^{\infty} \frac{3n - 2}{n!} = 3e - (2e - 2) = 3e - 2e + 2 = e + 2
$$

Thus, the sum of the series is $e + 2$, which corresponds to Option 2.

**Explanation for Other Options:**

1. **Option 1 (e):** This option suggests that the series converges to $e$. However, as shown in the breakdown, the series actually converges to $e + 2$, not just $e$.

2. **Option 3 ($e+3$):** This option suggests that the series converges to $e + 3$. The calculations clearly show that the additional constant term is 2, not 3.

3. **Option 4 ($e+4$):** This option suggests that the series converges to $e + 4$. Again, the calculations indicate that the correct additional constant is 2, not 4.

Therefore, based on the detailed breakdown and calculations, Option 2 ($e+2$) is indeed the correct answer.