The sum of the first n odd natural numbers is equal to n². For example, 1 = 1², 1+3 = 4 = 2², 1+3+5 = 9 = 3², and so on.

Substitute the simplified numerator back into the general term. We can simplify n²/n! by writing n! = n × (n-1)!, which allows us to cancel one n from the numerator and denominator.

The sum $1 + \tfrac{1+3}{2!} + \tfrac{1+3+5}{3!} + \tfrac{1+3+5+7}{4!} + \dots$ up to infinitely many terms is equal to:

Answer: $2e$

Step-by-step solution

Step 1: Identify the general term

First, we need to find a general expression for the $n$ -th term of the series. The numerator is the sum of the first $n$ odd natural numbers, and the denominator is $n!$ .

Step 2: Simplify the numerator

The sum of the first $n$ odd natural numbers is equal to $n^2$ . For example, $1 = 1^2$ , $1+3 = 4 = 2^2$ , $1+3+5 = 9 = 3^2$ , and so on.

Step 3: Rewrite the general term

Substitute the simplified numerator back into the general term. We can simplify $n^2/n!$ by writing $n! = n \times (n-1)!$ , which allows us to cancel one $n$ from the numerator and denominator.

Step 4: Split the general term

To make the terms suitable for summation, we split the numerator $n$ into $(n-1)+1$ . This allows us to further simplify the expression. Note that for $n=1$ , the term $1/(n-2)!$ is not defined, so we handle the first term separately.

Step 5: Calculate the sum of the series

The first term of the series is $T_1 = 1$ . For $n \ge 2$ , we use the simplified general term. The sum can be written as $T_1$ plus the sum of the split terms from $n=2$ to infinity. We recognize that $\sum_{k=0}^{\infty} \frac{1}{k!} = e$ (Euler's number). The second sum $\sum_{k=1}^{\infty} \frac{1}{k!}$ is $e-1$ . Combining these, we get the total sum.

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The sum 1+1+32!+1+3+53!+1+3+5+74!+…1 + \tfrac{1+3}{2!} + \tfrac{1+3+5}{3!} + \tfrac{1+3+5+7}{4!} + \dots1+2!1+3​+3!1+3+5​+4!1+3+5+7​+… up to infinitely many terms is equal to: