Solving Summation Using Induction: 1/((k-1)!(k+1)) < 1/k!

[topengonzo]

summation from k to n of 1/((k-1)!(k+1)) < 1/k! using induction

How can i write it to look math

 

[mathman]

You have 2 integer variables, k and n. What is the induction variable?

 

[topengonzo]

You have 2 integer variables, k and n. What is the induction variable?

n is induction variable and base term is n=k

 

[Fredrik]

When you said "summation from k to n", I assume you meant "summation from 1 to n". For all induction problems, keep in mind that the goal is to prove infinitely many statements with a finite amount of work. If we for each positive integer n denote the nth statement by P(n), the goal is to prove P(1), P(2), P(3) and so on, by only proving the following statements:

P(1)
For all positive integers j, if P(j), then P(j+1).

 

[Norwegian]

I assume the summation is from j to n, and that it ends with < 1/j! . It is not necessary to state the induction variable, but it is the lower limit j. It is btw much easier to just compute the sum without any induction nonsense.

 

[D H]

summation from k to n of 1/((k-1)!(k+1)) < 1/k! using induction

How can i write it to look math

This is how I would interpret what you wrote:
\sum_{r=k}^n \frac 1 {(k-1)!(k+1)} &lt; \frac 1 {k!}
This is "the sum from k to n of 1/((k-1)!(k+1))" -- and it is false.This is what I think you meant:
\sum_{r=k}^n \frac 1 {(r-1)!(r+1)} &lt; \frac 1 {k!}
This is true for all finite integers k>0, n≥k. You don't need recursion to prove this.

 

[topengonzo]

This is how I would interpret what you wrote:
∑r=kn1(k−1)!(k+1)<1k!

This is "the sum from k to n of 1/((k-1)!(k+1))" -- and it is false.This is what I think you meant:
∑r=kn1(r−1)!(r+1)<1k!

This is true for all finite integers k>0, n≥k. You don't need recursion to prove this.

Yes the second one you wrote is what i exactly mean. How do I prove it? Also I think if i set n -> infinity (find lim at infinity), I would get = instead of < . Am I correct?

 

[D H]

Yes the second one you wrote is what i exactly mean. How do I prove it? Also I think if i set n -> infinity (find lim at infinity), I would get = instead of < . Am I correct?

This is the homework section of PhysicsForums. You need to show some work.

I will give a hint: Induction is not the way to go here. Simply find the value of the series.

 

[topengonzo]

This is the homework section of PhysicsForums. You need to show some work.

I will give a hint: Induction is not the way to go here. Simply find the value of the series.

This is not the question of my homework. I will write it here down to show you that I am doing an effort and what I am asking will lead me to the answer.

Question:
Prove every positive rational number x can be expressed in ONE way in the form
x= a1 + a2/2! + a3/3! + ... + ak/k!
where a1,a2,...,ak are integers and 0<=a1,0<=a2<2,...,0<=ak<k
My solution:
To prove that there is only 1 solution, I am trying to prove
min(ak/k!)>max( (a(k+1))/(k+1)!+(a(k+2))/(k+2)!+...+ (a(k+n))/(k+n)! )
To get min set ak=1
and to get max set ak=k

and I get to the original question I am asking

 

[D H]

I'll give you a start. We agree that you are trying to prove
\sum_{r=k}^n \frac 1 {(r-1)!(r+1)} &lt; \frac 1 {k!}
Multiplying each term of the sum by one won't change anything, but write one as r/r.

 

[topengonzo]

I think that your previous answer using series will solve it.

I am thinking of putting

\sum_{r=k}^n \frac r {(r+1)!} &lt; \frac 1 {k!}



and I use the poisson distribution series to prove = 1/k! and thus the sum of any finite terms of the series is < 1/k!

Am I correct?

 

[D H]

You're thinking too hard. Just rewrite the series so the sum goes from k+1 to n+1.

 

[topengonzo]

\sum_{r=k+1}^{n+1} \frac {r-1} {(r)!} &lt; \frac 1 {k!}
Is this correct?
\sum_{r=0}^{inf} \frac {r-1} {(r)!} = e-e=0

And then I take out term from 0 to k?

 

[D H]

\sum_{r=k+1}^{n+1} \frac {r-1} {(r)!} &lt; \frac 1 {k!}
Is this correct?

Yes!

\sum_{r=0}^{inf} \frac {r-1} {(r)!} = e-e=0

And then I take out term from 0 to k?

No!

You are thinking too hard! Look at the numerator. Rewrite that finite sum as the difference between two sums. Those two sums will have n-1 common terms. Get rid of them.

 

[topengonzo]

\sum_{r=k+1}^{n+1} \frac {1} {(r-1)!} - \sum_{r=k+1}^{n+1} \frac {1} {(r)!}

YOUR RIGHT! it becomes simply
\frac {1} {(k)!} - \frac {1} {(n+1)!}

Just 1 final question?

Prove every positive rational number x can be expressed in ONE way in the form
x= a1 + a2/2! + a3/3! + ... + ak/k!
where a1,a2,...,ak are integers and 0<=a1,0<=a2<2,...,0<=ak<k

I take lim as n tends to infinity of my answer so I get that the infinite sum of all terms after ak/k! is equal to ak/k! which means there is no overlapping and one and only one way to represent x.
Also the sum of fractions span from 0 to infinity which means EVERY positive rational number x can be represented by the sum of fractions

 

[topengonzo]

Is my solution correct and is there another way to solve it?