Prove (1/2!) + (2/3!) + ……….. +(n/(n+1)!) = 1-1/(n+1)!

[solomon_13000]

Prove (1/2!) + (2/3!) + ……….. +(n/(n+1)!) = 1-1/(n+1)!

Prove the following proposition for n > 1 :

(1/2!) + (2/3!) + ……….. +(n/(n+1)!) = 1-1/(n+1)!

my solution:

Assume n = p
The next step n = p + 1

[(1/2!) + (2/3!) + ……….. +(p/(p+1)!)] +((p+1)/((p+1)+1)!) = 1-1/((p+1)+1)!

1-1/(p+1)! + ((p+1)/(p+2)!)

I am confuse with solving the problem using !.

How do I solve the problem.

 

[pka]

\(\displaystyle \L\begin{array}{rcl}
1 - \frac{1}{{\left( {p + 1} \right)!}} + \frac{{\left( {p + 1} \right)}}{{\left( {p + 2} \right)!}} & = & 1 + \frac{{ - \left( {p + 2} \right) + \left( {p + 1} \right)}}{{\left( {p + 2} \right)!}} \\
& = & 1 - \frac{1}{{\left[ {\left( {p + 1} \right) + 1} \right]!}} \\
\end{array}\)

 

[stapel]

my solution: Assume n = p
The next step n = p + 1

So you're still not doing the base step (in this case, for n = 2)?

And you'll never be able to do the "n = p + 1" step until you start doing the "n = p" step correctly. You're not supposed to "assume that n = p", and then go off to do the next step. You're supposed to "assume that the proposition is true for n = p", and make the meaning of that assumption clear, so you can then use this assumption in the "n = p + 1" step.

You don't seem to be understanding this from our many replies. Please talk with your instructor about this, because you really do need to understand what is going on.

Thank you.

Eliz.

 

[pka]

So you're still not doing the base step (in this case, for n = 2)?

Yes, I thought that also when I first read it.
But the base number is 1 not 2. Look at the numerators.
I give you that it is true for n=1.

 

[stapel]

So you're still not doing the base step (in this case, for n = 2)?

I give you that it is true for n=1.

But the exercise specifies to prove the proposition for "n > 1". Regardless of where all the proposition might work, the student would appear to be expected to start at "n = 2".

I could be wrong, of course....

Eliz.

 

[pka]

But the exercise specifies to prove the proposition for "n > 1". Regardless of where all the proposition might work, the student would appear to be expected to start at "n = 2".

I think that the student was into the proof, at the final induction stage, where n is greater that 1.