Question

What is the value of `sum_(k = 0)^oo k^2/(k!)`?

Answer 1

I'd use the ratio test to test for convergence.

`L = lim_(k->oo) |((k + 1)^2/((k + 1)!))/(k^2/(k!))|`

`L = lim_(k->oo) |(k^2 + 2k + 1)/((k + 1)!) * (k!)/k^2|`

`L = lim_(k->oo) |(k^2 + 2k + 1)/((k + 1)k!) * (k!)/k^2|`

`L = lim_(k->oo) |((k + 1)(k + 1))/((k^2)(k + 1))|`

`L = lim_(k->oo) |(k + 1)/k^2|`

`L = lim_(k->oo) |k/k^2 + 1/k^2|`

`L = lim_(k->oo) |1/k + 1/k^2|`

`L = 0 + 0`

`L = 0`

Since this is smaller than `1`, the series must converge to some finite number. However, finding the sum seems to be a little bit more troublesome. According to the computer, the sum is `2e`, but how you would prove this is beyond me.

I'll leave this to other, more clever contributors :)

Hopefully this helps!

Answer 2

`2e`

Explanation:

Knowing that

`e^x = sum_(k=0)^oo x^k/(k!)` and that

`e = sum_(k=0)^oo 1/(k!)` we have

`sum_(k=0)^oo k^2/(k!) = sum_(k=1)^ook/((k-1)!) = sum_(k=0)^oo (k+1)/(k!) = sum_(k=1)^oo 1/((k-1)!)+sum_(k=0)^oo 1/(k!) = 2 e`

Answer 3

`S =sum_"k=0"^oo k^2/(k!) =2e`

Explanation:

@HSBC244 has already proved that the sum converges. This is a good start and no need to reproduce here,

So now let's consider the terms of `S`

`S =sum_"k=0"^oo k^2/(k!) = sum_"k=0"^oo k/((k-1)!)`

`:. S = 0 +1/(0!) +2/(1!) + 3/(2!) + 4/(3!) + ......`

`= 1+2/(1!)+3/(2!)+4/(4!)+ .........`

`= 1+ (1+1)/(1!) + (1+2)/(2!) + (1+3)/(3!) + ......`

We can now split `S` into two sums as follows:

`S_1 = 1+1/(1!) + 1/(2!) + 1/(3!) + 1/(4!)+ .....`
Plus
`S_2 = 1/(1!) + 2/(2!) + 3/(3!) + 4/(4!) ........`

Notice that `S_1` is the series expansion of `e`

Now analysing `S_2`

`S_2 = 1/(1!) + 2/(2!) + 3/(3!) + 4/(4!) ........`

`= 1 + 1/(1!) + 1/(2!) + 1/(3!) + ........`

Which again is the series expansion of `e`

So, Since `S = S_1+S_2`

`S= e+e = 2e`