# Can you show by induction that `sum_(k=1)^n k/((k+1)!)= ((n+1)!-1)/((n+1)!)` where n is a natural number.This is certainly not very difficult.

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To prove by induction, you just have to prove that two things are true:

1) The base case (for us, the case where n = 1)

2) The case in which `n=a+1` given that the equation holds for `n=a`

``So let's go ahead and take care of the base case:

If n = 1, then our sum becomes:

`sum_(k=1)^1k/((k+1)!)=1/(2!) = 1/2`

If we check the other side of the equation:

`((1+1)!-1)/((1+1)!)=(2-1)/2 = 1/2`

Therefore, our base case holds. Now, we can proceed to the inductive step:

Now, we'll assume that the following is true (when `n = a` ):

`sum_(k=1)^a k/((k+1)!)=((a+1)!-1)/((a+1)!)`

Let's prove that the same equation holds for when `n=a+1`. Let's start with the sum:

`sum_(k=1)^(a+1)k/((k+1)!)`

Now, recognize that this shorthand can split off the last term in the following way:

`=sum_(k=1)^a k/((k+1)!) + (a+1)/((a+2)!)`

Now, we recognize part of our assumption. Let's go ahead and apply our assumption to substitute for the summation term:

`=((a+1)!-1)/((a+1)!) + (a+1)/((a+2)!)`

Now, to add two fractions, we must have like denominators, so we'll multiply the term of the left by `(a+2)/(a+2)` (recall, x! * (x+1) = (x+1)!):

`((a+1)!(a+2) - (a+2))/((a+1)!(a+2)) = ((a+2)! - (a+2))/((a+2)!)`

Now, let's substitute this for the term of the left two equations above:

`= ((a+2)!-(a+2))/((a+2)!) + (a+1)/((a+2)!)`

We can combine the numberators now:

`((a+2)!-a-2+a+1)/((a+2)!) = ((a+2)! - 1)/((a+2)!)`

And that completes our proof! We just showed that if the equation holds true for `n=a` then the equation must hold for `n=a+1`. Combining this fact with our proof that the base case holds true means that we have proven the equation holds true for all `ninNN`.

Hope that helps! See the link below for more info on induction. ``

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