The sum 1+2!1+3+3!1+3+5+4!1+3+5+7+… up to infinitely many terms is equal to:
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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 n2. For example, 1=12, 1+3=4=22, 1+3+5=9=32, and so on.
Step 3: Rewrite the general term
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.
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 T1=1. For n≥2, we use the simplified general term. The sum can be written as T1 plus the sum of the split terms from n=2 to infinity. We recognize that ∑k=0∞k1!=e (Euler's number). The second sum ∑k=1∞k1! is e−1. Combining these, we get the total sum.