# Homework Help: Finding value

1. Dec 29, 2014

### Raghav Gupta

1. The problem statement, all variables and given/known data

Let $S_k , k = 1,2,3,…,100$ denote the sum of the infinite geometric series whose first term is $\frac{k-1}{k!}$ and the common ratio is $\frac {1}{k}$. Then value of $\frac {100^2}{100!} + \sum\limits_{k=1}^{100} | (k^2 - 3k + 1)S_k |$ is
2. Relevant equations
Sum of infinite geometric series is a/(1-r) where a is first term and r the common ratio.

3. The attempt at a solution
Got $S_k= \frac{1}{(k-1)!}$

2. Dec 29, 2014

### BvU

And: S1 = 1, even if all terms are 0 ?

3. Dec 29, 2014

### Raghav Gupta

Yeah, the problem statement is correct. 100! Is a great number indeed but it may reduce I think so by solving.
Didn't understand And: S1 = 1, even if all terms are 0 ?

4. Dec 29, 2014

### BvU

S1: first term (1-1)/1! all other terms: (1-1) / (1!*1n) n = 2,3,4...

whereas 1/(1-1)! = 1

Last edited: Dec 29, 2014
5. Dec 29, 2014

### Raghav Gupta

You are right,
So, $S_k=\frac{1}{(k-1)!}$ is relevant for k >= 2 for k=1 the sum is 0.

6. Dec 30, 2014

### Raghav Gupta

That's true but that doesn't mean my question is wrong.What about next?
We have got Sk as 0 for k=1
Well the solution to the problem is 3, if it makes one easy to solve.
But I don't know how to get it.

Last edited: Dec 30, 2014
7. Dec 30, 2014

### Dick

The sum is 3. I did it in maxima. It must telescope in some clever way. But I still don't see how.

Last edited: Dec 30, 2014
8. Dec 30, 2014

### BvU

Makes two [edit:] three of us. I can cheat with a spreadsheet and see that even after 11 terms the summation is already at 3 to within 6 significant digits (using Sk = 0 for k = 1), but that's probably not the idea for this exercise.

Especially the presence of this 1002/100! is intriguing: if the correct, exact answer is a Taylor series, that would mean that this expression is the $\displaystyle \sum_{k=101}^\infty$ (with a value of about 1.07 * 10-154 ) . But a Taylor series of what function ? I can't find out.

9. Dec 30, 2014

### Dick

I don't think it's a Taylor series. It's some clever way of expressing f(k)=(k^2-3k+1)/(k-1)! as f(k)=g(k)-g(k-1) so it telescopes. What's g(k)?

10. Dec 30, 2014

### haruspex

Seems fairly straightforward. First, sum the infinite series to get Sk into closed form.
You can get rid of the modulus signs by recognising which few terms would be negative otherwise.
Then use the closed form of Sk to simplify the expression in the finite sum.
As Dick suspected, you get cancellation of all but the first few and last few terms.

11. Dec 30, 2014

### haruspex

My post crossed with yours. Don't assume it'll be g(k-1).

12. Dec 30, 2014

### Dick

Well, it's 5AM here. g(k-2)?

13. Dec 30, 2014

### haruspex

Yes (10pm).

14. Dec 30, 2014

### Dick

Thanks. That was what was eluding me.

15. Dec 30, 2014

### Raghav Gupta

I am a high school student. So Taylor series questions are not usually given to that advanced level.
So something other might be applied here.
Sorry to both of you as I can't figure out how the cancellation is taking place.
Well if one is finding it difficult to explain by latex ( as much time is consumed in that ). He/She can write in a paper and upload that as this question is very important for me.
Please if one is not able to figure out the solution can he/she ask his/her friend/neighbor/ or anybody.
Well this might be also my last question for PF and any further help would be appreciated.
To some people it might be seeming why I am giving so reasons for the sake of this single question.
But it is really important for me.
Hope one might understand

16. Dec 30, 2014

### BvU

Well, as Dick and Haru indicated, I was thinking in the wrong direction. So forget Taylor and think what a summation ∑ f(k) will do if term f(k) can be written as a difference g(k) - g(k-2) ... a lot of stuff cancels (the 'telescope' description)

Just so you know: to D&H it seems to be obvious, but I still don't have a clue what g(k) could be !

17. Dec 30, 2014

### Dick

Here's kind of a large hint. $$k^2-3k+1=(k-2)(k-1)-1$$.

18. Dec 30, 2014

### Raghav Gupta

Well this was the masterpiece for solving the question. How did you think of it? I got the answer. The rest work was of manipulation. I am not posting my solving method as it would require a lot of latex. Thanks Dick. I would like to thanks others also for replying (BvU, haruspex). This also proved to me for the time being at high school that PF could solve any type of question. It's a matter of understanding. Thanks all.

Last edited: Dec 30, 2014
19. Dec 30, 2014

### Dick

I thought of it because if I see a finite series that isn't arithmetic or geometric that has a nice sum then it must telescope somehow. I.e. I can rewrite the summand in such a way that there will be a lot of cancellation. I was being a little thick about how to do it until haruspex gave me a nudge.

20. Dec 30, 2014

### Raghav Gupta

:)
This might have given you the nudge.

21. Dec 30, 2014

### Dick

This was the 'nudge'.

22. Dec 30, 2014

### Raghav Gupta

Okay.
It is amazing sometimes that without posting big latex, one figures out solution. Isn't it?

23. Dec 30, 2014

### haruspex

Further thought:
Show that if we replace the quadratic in the question by some other polynomial P(k), degree n, then such a collapse can be found if and only if $\Sigma _{i=0}^n \frac{P(i)}{i!} \Sigma _{j=0}^{n-i} \frac{(-1)^j}{j!} =0$.
Btw, seems like Dick and I arrived at the solution by slightly different routes. Dick explicitly looked for cancelling pairs (right?). I looked at the quadratic/factorial expression and treated it like a partial fractions problem. First extract the remainder modulo k-1 to get P(k)(k-1)+constant, giving constant/(k-1)! + P(k)/(k-2)!, then reduce the second term in the same way wrt k-2. That way you can find reductions involving any number of 'g()' terms. For collapse to occur, the sum of the constants must turn out to be zero.