Riemann Sum approximation help?

Question

It's the sum from K=1 to infinity for (K)/(K!)

It's merely auxillary to another problem (statistics.....poisson distribution).....

But I've forgotten how to do this one, and I don't have a book on integrals and sums with me.

Answer 1

k/k! = 1/(k-1)!

Formally replace k by j+1; then the sum becomes

sum from j=0 to infinity of 1/j!

Recognize it?

Answer 2

The sum of K=1 to infinity for (K)/(K!) is equal to e.
It's easy to verify this if you remember that e can be expressed as the limit of (1+1/n)^n when n tends to infinity.
Or you may also get this if you remember the Taylor expansion of the function e^x (which is 1+x+x^2/2!+x^3/3!+...) and just substitute x=1.

Hope this helps!

Answer 3

Σ [k=1.. ∞] k/k!
= Σ [k=1.. ∞] 1/(k-1)!
= Σ [m=0.. ∞] 1/(m!)
= e.


Note: e^x = Σ [j=0.. ∞] x^j/(j!)

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Answer 4

it's e