To solve the given series:

$S = \sum_{n=0}^{\infty} \frac{5n + 1}{(2n + 1)!}$

we will approach the problem step by step.

Step 1: Break the series into two separate sums

The given expression involves both $5n$ and $1$ in the numerator. We can split the series into two sums:

$S = \sum_{n=0}^{\infty} \frac{5n}{(2n+1)!} + \sum_{n=0}^{\infty} \frac{1}{(2n+1)!}$

Thus, we have two sums to compute:

1. $S_1 = \sum_{n=0}^{\infty} \frac{5n}{(2n+1)!}$
2. $S_2 = \sum_{n=0}^{\infty} \frac{1}{(2n+1)!}$

Step 2: Solve for $S_2$

The second sum, $S_2$, is a known series. This is related to the expansion of $\sinh(x)$, which is given by:

$\sinh(x) = \sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!}$

For $x = 1$, we get:

$\sinh(1) = \sum_{n=0}^{\infty} \frac{1^{2n+1}}{(2n+1)!} = \sum_{n=0}^{\infty} \frac{1}{(2n+1)!}$

Thus,

$S_2 = \sinh(1)$

Using the approximation $\sinh(1) \approx 1.175201$, but for symbolic accuracy, we'll keep it in the form $\sinh(1)$.

Step 3: Solve for $S_1$

Now consider the sum $S_1 = \sum_{n=0}^{\infty} \frac{5n}{(2n+1)!}$.

This sum can be rewritten as:

$S_1 = 5 \sum_{n=0}^{\infty} \frac{n}{(2n+1)!}$

We can split $n$ as:

$n = \frac{(2n+1) - 1}{2}$

Thus,

$S_1 = \frac{5}{2} \sum_{n=0}^{\infty} \frac{(2n+1)}{(2n+1)!} - \frac{5}{2} \sum_{n=0}^{\infty} \frac{1}{(2n+1)!}$

The first term simplifies to:

$\sum_{n=0}^{\infty} \frac{(2n+1)}{(2n+1)!} = \sum_{n=0}^{\infty} \frac{1}{(2n)!} = \cosh(1)$

where $\cosh(x)$ is the hyperbolic cosine. For $x=1$, $\cosh(1)$ is approximately $1.54308063482$.

Thus, $S_1$ becomes:

$S_1 = \frac{5}{2} \left( \cosh(1) - \sinh(1) \right)$

Using the known values of $\cosh(1) \approx 1.54308063482$ and $\sinh(1) \approx 1.17520119364$, we get:

$S_1 \approx \frac{5}{2} (1.54308063482 - 1.17520119364) = \frac{5}{2} \times 0.36787944118 \approx 0.91969860295$

Step 4: Combine $S_1$ and $S_2$

Now we can combine the two sums to find the total:

$S \approx 0.91969860295 + 1.17520119364 = 2.09489979659$

Step 5: Relating to the known result

The series is known to be equal to:

$S = \frac{e}{2} + \frac{2}{e}$

Using $e \approx 2.71828$, we calculate:

$\frac{e}{2} + \frac{2}{e} = \frac{2.71828}{2} + \frac{2}{2.71828} \approx 1.35914 + 0.73576 = 2.0949$

Thus, the series indeed evaluates to:

$S = \frac{e}{2} + \frac{2}{e}$

Conclusion

The sum of the given infinite series is:

$\boxed{\frac{e}{2} + \frac{2}{e}}$

Would you like further clarification or any additional details?

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Here are 5 related questions:

1. What are some common properties of hyperbolic functions like $\sinh(x)$ and $\cosh(x)$?
2. How can Taylor series expansions be used to approximate functions?
3. What are some other famous series related to $e$?
4. How is the series expansion of $\sin(x)$ different from $\sinh(x)$?
5. What are the convergence criteria for infinite series?

Tip: Understanding hyperbolic functions is helpful in solving many types of series, as they often appear in physics and engineering applications.