Euler limit | Problem based on Euler's number

Let $ \{a_n\}_{n\ge 1} $ be a sequence of real numbers such that $$ a_n = \frac{1 + 2 + … + (2n-1)}{n!} , n \ge 1 $$ . Then $ \sum_{n \ge 1 } a_n $ converges to ____________

Notice that $ 1 + 3 + 5 + … + (2n-1) = n^2 $. A quick way to remember this is sum of first n odd numbers is $ n^2 $ Hence $ a_n = \frac{n^2}{n!} $ Simplifying this a little we have $ a_n = \frac {n}{(n-1)!} $

Notice that n = n-1 +1 Then $ a_n = \frac { n-1}{(n-1)!} + \frac {1}{ (n-1)!} = \frac { 1}{(n-2)!} + \frac {1}{ (n-1)!} $ We are interested in the limit of the partial sums of $a_n $

Recall the $ \sum \frac {1}{n!} $ converges to the Euler number e. (Do not know this? Look up the definition of e in the internet). Here we have two copies of this series if we sum $ a_n $. (Since we are taking limit of a sum, (n-1) and (n-2) does not matter). Hence the limit of this sum is 2e,