Sum of a Miscelleneous series.

1. Oct 22, 2011

AGNuke

If $f(r) = 1 + \frac{1}{2} + \frac{1}{3}+... \frac{1}{r}$ and $f(0) = 0$, then find $\sum_{r=1}^{n}(2r+1)f(r)$

I tried the following steps but stuck at some point.

$\sum_{r=1}^{n}rf(r)+(r+1)f(r)$

$= \sum_{r=1}^{n}rf(r)+(r+1)[f(r+1)-\frac{1}{r+1}]$

$= \sum_{r=1}^{n}rf(r)+(r+1)f(r+1) - \sum_{r=1}^{n}1$

The real problem is to find $\sum_{r=1}^{n}rf(r)+(r+1)f(r+1)$

For verification, the Solution is $(n+1)^{2}f(n+1)-\frac{n^2+3n+2}{2}$

Last edited: Oct 22, 2011
2. Oct 22, 2011

Staff: Mentor

Then $\sum_{r=1}^{n}(2r+1)f(r)$ does what?
Don't use $\Rightarrow$ here. Connect expressions that are equal with =.

3. Oct 22, 2011

AGNuke

Fixed

Fixed

4. Oct 23, 2011

AGNuke

Anyone even with the slightest idea what to do next? I don't have much time left to submit the solution and I am also working on the solution simultaneously.

Thanks!

5. Oct 23, 2011

I like Serena

Hi AGNuke!

I do not know how to derive your solution, but you can prove it with full induction.
Would full induction satisfy the requirements?

6. Oct 23, 2011

AGNuke

It's a problem of Summation for a product of two functions of a natural number. One is quite easy but the given f(r) is quite complex to sum.

I am like in 11th grade or something, so I really don't know much about advanced mathematics.

7. Oct 23, 2011

I like Serena

I'd classify any derivation or proof as advanced mathematics.

Without it I can only recommend trial and error.
If you have multiple answers you can verify the answers by trying them for n=1 and n=2.
And if that doesn't leave one answer, try n=3.

8. Oct 23, 2011

AGNuke

That's what we call objective approach. The answer is satisfying n=1. But will it not be good if I can do the same with the proper subjective approach?

9. Oct 23, 2011

I like Serena

I'm not sure what you mean by subjective approach, but I do not see another approach that you can complete within 3 minutes.

10. Oct 23, 2011

AGNuke

Subjective approach is pure mathematics + Some tricks based on assumptions and previous knowledge to cut down the time.

I know this question is ultra level, that's why we weren't able to solve it. But since we've got plenty of time after the paper, I was considering that I can get a proper subjective approach on papers.

11. Oct 23, 2011

I like Serena

Well, I guess I can give you an estimation method.

We have these things called the order of a sequence or series.
For instance 2n+1 has order n, and n^2 + 3 has order n^2.
The series $\sum\limits_r 2r+1$ has order n^2, since it's an arithmetic progression.

Since f(r) is a diverging series, but diverges slower than r, it has order n.

With these you can determine the order of your original expression and check what the order of your answers is.

12. Oct 23, 2011

AGNuke

Ok... Thanks for this method, but anyways, I got my answer after an hour or so of paperwork. I'll be posting it shortly to see where could I cut through steps and time.

13. Oct 24, 2011

AGNuke

Here's my attempt at the question. Got my answer but still I feel it is quite unorthodox. I also want to see where can I cut short some steps.

We know that and expanding Ʃrf(r)+Ʃ(r+1)f(r+1), we get

As we can see, 1f(1) is simply equals to 1.

Expanding rf(r) so as to find something more useful.

If we group them properly, what we see, we see some individual series with a common denominator, and what's more, they are in AP! And taking out their sum is no issue.

Sum of AP in the first bracket is

Sum of AP in the second bracket is

Sum of AP in the third bracket is

and if we continue to follow the trend, we get the sum of the last bracket as

Here's my mischievous trick : If I consider another variable term k, which is the order of the bracket : first, second, third,..., nth bracket as k = 1, 2, 3,..., n. This forms a series of sum of sum of different APs, and the term of the main series is given by:

By doing this dirty work, I converted the summation of seemingly-impossible-to-sum series into a nicer series. Hence,

Using this equation into long forgotten step:

Expanding our new summation function, we get

(Sorry for eating out intermediate step, was too lazy to type that out)

By observation, we see ; and

By further manipulation,

Therefore, the next step is

Grouping common terms:

Tada! Here comes the answer but at this cost... The thing I am asking for is How to cut short through the procedure via a better trick or a concept (not too high level, I'm only in 11th grade)

Last edited: Oct 24, 2011
14. Oct 24, 2011

I like Serena

Congratulations! Looks good!!

I see you've already taken the time to make it as readable as possible.
The bad news: I see no possible shortcuts.

15. Oct 24, 2011

AGNuke

What's the use of a solution if even I can't read it.

Sorry to hear that you can't see any possible shortcuts. Anybody else?