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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,

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