Is there any general formula for

We do know that f(1) equals:
1+1/1 +1/2! +1/3!=e (the constant)
so f(1) =e
f(0)=infinity (1+1+1+1+1+1+1+1......)
now, we have to see how it changes from f(1) to f(2) and so on....
f(2) = 1 + 1/k! (k only even numbers)
f(3) = 1 + 1/s! (s only numbers the 3 divides them)

now, is there some genereralization???

Brad,

I assume x is an integer. Set w=exp(2ip/x). Now try expanding:

1+e^w+e^w2+¼+e^wx-1

(using the exponential power series), and see what you get...

I think I'm wrong on this, but I get x+1-e. I can show my work if need be...

Brad

Note that it is:

x-1 å m=0

ewm=

x-1 å m=0

¥ å n=0

wm n/n! =

¥ å n=0

x-1 å m=0

wm n/n!=

¥ å n=0

1/n!× [1+wn+w2n+¼+w(x-1)n]

The term in square brackets is obviously x if wⁿ=1, i.e. if n is a multiple of x; otherwise it's 0 (using theformula for a geometric progression). So the expression is:

¥ å n=0

1/n!×x

where the sum is now only over values of n which are multiples of x. Therefore it works out as x(1/0!+1/x!+1/(2x)!+¼) which is x times your original sum. So your original sum is:

1/x×

x-1 å m=0

ewm

(*)

We can write this in terms of trig functions, using the fact that w^m=exp(2pi m/x)=cos(2pm/x)+isin(2pm/x) we get your sum to be:

1/x×

x-1 å m=0

ecos(2pm/x)+isin(2pm/x)

=1/x×

x-1 å m=0

ecos(2pm/x)(cos(sin(2pm/x))+isin( sin(2pm/x)))

But since the answer is real, the imaginary part must vanish.

[Which means that we have proven, as a bonus, that

x-1 å m=0

ecos(2pm/x)sin(sin(2pm/x))=0

]

So your sum works out as:

1/x×

x-1 å m=0

ecos(2pm/x) cos(sin(2pm/x))

I'm not sure that's helpful. Quite possibly (*) is more useful than our final answer. At any rate at least it's a finite sum now:-)

Wow michael,
how did you even think of this !!!

love arun