infinite series

Question

the sum of the series

1/1! + (1+2)/ 2! + (1+2+3)/3! +.......... equals

 

a) e             b) e/2                c)3e/2                  d) 1+ e/2

 

please help mesir, i did try my best but i couldnt 

Answer 1

First of all, notice that the numerator of the nth term is the series 1+2+3+...+n.  That's a simple arithmetic series that can be expressed as n(n+1)/2.

 

So the original series can be written, using summation notation, as:  ₁∑^∞ (n(n+1)/2) / n! = 1/2 ₁∑^∞ n(n+1) / n!

 

You can cancel the n in the numerator and denominator and it becomes 1/2 ₁∑^∞ (n+1) / (n-1)!

 

Now we do a trick of changing the bottom limit of the summation to 0.  To do that we have to change each n to n+1.  So we get 1/2 ₀∑^∞ (n+2) / n!

 

Separate this into two sums:  1/2 ₀∑^∞ n / n! + 1/2 ₀∑^∞ 2 / n!

 

In the first sum, we'd like to cancel the n from the numerator and denominator again, and it becomes 1/2 ₀∑^∞ 1 / (n-1).  But that doesn't make sense because the first term of the sum would have a denominator of (-1)!.  So lets take the first term out of the sum before we cancel:  1/2 (0/0!) + 1/2 ₁∑^∞ n / n!.  But that term we removed is just 0, so we can eliminate it.  So the first sum is equal to 1/2 ₁∑^∞ n / n! and after cancelling n, it is 1/2 ₁∑^∞  1 / (n-1)!.

 

Let's do the trick of changing the bottom limit of the summation to 0 again:  1/2 ₁∑^∞  1 / (n-1)! = 1/2 ₀∑^∞  1 / n!.  We can recognize this as 1/2 e = e/2.

 

Now let's go back to the second sum we had after we did the separation -- 1/2 ₀∑^∞ 2 / n!.  This is (1/2)(2) ₀∑^∞ 1 / n! = ₀∑^∞ 1 / n! which we can again recognize as e. 

 

So the two sums together are e/2 + e = 3e/2.  So the answer is (c).