Σ [(5n + 1) / (2n + 1)!] 
where n goes from 0 to infinity 
= e/2 + 2/e

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.

Σ [(5n + 1) / (2n + 1)!] where n goes from 0 to infinity = e/2 + 2/e

Explore the detailed step-by-step solution of the infinite series Σ [(5n + 1) / (2n + 1)!], demonstrating its evaluation to e/2 + 2/e using hyperbolic function expansions such as sinh(x) and cosh(x). Suitable for advanced students in undergraduate mathematics.

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