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,## Get Started with I.S.I. Entrance Program

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