Question

Let \( \{a_n\}_{n \geq 1} \) be a sequence of real numbers such that \[ a_n = \frac{1 + 3 + 5 + \dots + (2n - 1)}{n!}, n \geq 1. \] Then \( \sum_{n \geq 1} a_n \) converges to ................

Hint:

For series with factorials in the denominator, use the ratio test to determine convergence. Factorials grow very fast, which usually leads to convergence.

Answer 1

Correct Answer: 5.4 - 5.5

Step 1: Simplify the numerator

The sum of the first (n) odd numbers is
\[1+3+5+\cdots+(2n-1)=n^2.\]
Hence
\[a_n=\frac{n^2}{n!}.\]

Step 2: Rewrite \(n^2\)

\[n^2=n(n-1)+n.\]
Thus
\[\frac{n^2}{n!} =\frac{n(n-1)}{n!}+\frac{n}{n!} =\frac{1}{(n-2)!}+\frac{1}{(n-1)!}.\]

Step 3: Sum the series

\[\sum_{n=1}^\infty a_n =\sum_{n=1}^\infty \frac{1}{(n-2)!} +\sum_{n=1}^\infty \frac{1}{(n-1)!}.\]

Adjust indices:

For the first sum, terms start effectively from \(n=2\):
\[\sum_{n=1}^\infty \frac{1}{(n-2)!} =\sum_{k=0}^\infty \frac{1}{k!} =e.\]

For the second sum:
\[\sum_{n=1}^\infty \frac{1}{(n-1)!} =\sum_{k=0}^\infty \frac{1}{k!} =e.\]

Therefore,
\[\sum_{n=1}^\infty a_n = e+e = 2e.\]

Final answer

\[\boxed{\sum_{n=1}^\infty a_n = 2e \approx 5.436}\]