Sum of a Miscelleneous series.

In summary, In order to solve this problem, you need to use a subjective approach and determinate the order of the series.
  • #1
AGNuke
Gold Member
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If [itex]f(r) = 1 + \frac{1}{2} + \frac{1}{3}+... \frac{1}{r}[/itex] and [itex]f(0) = 0[/itex], then find [itex]\sum_{r=1}^{n}(2r+1)f(r)[/itex]

I tried the following steps but stuck at some point.

[itex]\sum_{r=1}^{n}rf(r)+(r+1)f(r)[/itex]

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

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

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

Can someone please help me beyond this point by solving the above steps?

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

Thanks in Advanced!
 
Last edited:
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  • #2
AGNuke said:
If [itex]f(r) = 1 + \frac{1}{2} + \frac{1}{3}+... \frac{1}{r}[/itex] and [itex]f(0) = 0[/itex], then [itex]\sum_{r=1}^{n}(2r+1)f(r)[/itex]
Then [itex]\sum_{r=1}^{n}(2r+1)f(r)[/itex] does what?
AGNuke said:
I tried the following steps but stuck at some point.

[itex]\sum_{r=1}^{n}rf(r)+(r+1)f(r)[/itex]

[itex]\Rightarrow \sum_{r=1}^{n}rf(r)+(r+1)[f(r+1)-\frac{1}{r+1}][/itex]
Don't use [itex]\Rightarrow[/itex] here. Connect expressions that are equal with =.
AGNuke said:
[itex]\Rightarrow \sum_{r=1}^{n}rf(r)+(r+1)f(r+1) - \sum_{r=1}^{n}1[/itex]

The real problem is to find [itex]\Rightarrow \sum_{r=1}^{n}rf(r)+(r+1)f(r+1)[/itex]

The Solution is [itex](n+1)^{2}f(n+1)-\frac{n^2+3n+2}{2}[/itex]

Thanks in Advanced!
 
  • #3
Mark44 said:
Then [itex]\sum_{r=1}^{n}(2r+1)f(r)[/itex] does what?
Fixed

Don't use [itex]\Rightarrow[/itex] here. Connect expressions that are equal with =.
Fixed
 
  • #4
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
Hi AGNuke! :smile:

I do not know how to derive your solution, but you can prove it with full induction.
Would full induction satisfy the requirements?
 
  • #6
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
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
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
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
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
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 [itex]\sum\limits_r 2r+1[/itex] 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
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. :biggrin:
 
  • #13
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.

gif.latex?f(r)=%201+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{r};%20\sum_{r=1}^{n}(2r+1)f(r).gif


gif.latex?=\sum%20[r+(r+1)]f(r).gif


gif.latex?=\sum%20rf(r)+\sum%20(r+1)f(r).gif


gif.latex?=\sum%20rf(r)+\sum%20(r+1)\left%20[%20f(r+1)-\frac{1}{(r+1)}%20\right%20].gif


gif.latex?=\sum%20rf(r)+\sum%20(r+1)f(r+1)-\sum%201.gif


We know that
gif.latex?\sum_{r=1}^{n}1=n.gif
and expanding Ʃrf(r)+Ʃ(r+1)f(r+1), we get

gif.latex?\dpi{120}%20\left%20[%201f(1)+2f(2)+3f(3)+...gif
gif.latex?\dpi{120}%20\left%20[2f(2)+3f(3)+...gif


gif.latex?\dpi{120}%20=\left%20[%201f(1)+2f(2)+3f(3)+...gif
gif.latex?\dpi{120}%20\left%20[1f(1)+2f(2)+3f(3)+...gif


gif.gif


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

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

gif.gif
gif.gif
gif.gif
gif.gif
gif.gif
gif.latex?\dpi{120}%20+...+\left%20(\frac{n}{1}+\frac{n}{2}+\frac{n}{3}+%20\frac{n}{4}+...gif


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.

gif.gif
gif.latex?\dpi{120}%20=\left%20(%20\frac{1}{1}+\frac{2}{1}+\frac{3}{1}+\frac{4}{1}+...gif
gif.latex?\dpi{120}%20+\left%20(%20\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+...gif
gif.latex?\dpi{120}%20+\left%20(%20\frac{3}{3}+\frac{4}{3}+...gif
gif.latex?\dpi{120}%20+...gif


Sum of AP in the first bracket is
gif.gif


Sum of AP in the second bracket is
gif.gif


Sum of AP in the third bracket is
gif.gif


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


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:

gif.gif


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

gif.gif


Using this equation into long forgotten step:

gif.gif


Expanding our new summation function, we get

gif.gif

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

By observation, we see
gif.gif
;
gif.gif
and
gif.gif


By further manipulation,

gif.gif


Therefore, the next step is

gif.gif


Grouping common terms:

gif.gif


gif.gif


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)

Hoping for a reply soon!
 
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  • #14
Congratulations! Looks good! :smile:

I see you've already taken the time to make it as readable as possible.
The bad news: I see no possible shortcuts.
 
  • #15
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?
 

FAQ: Sum of a Miscelleneous series.

What is a miscellaneous series?

A miscellaneous series is a mathematical series where the terms do not follow a specific pattern or formula. Each term is unique and unrelated to the previous terms.

How do you find the sum of a miscellaneous series?

The sum of a miscellaneous series can be found by adding up all the terms in the series. There is no specific formula or shortcut for finding the sum, so each term must be added individually.

What is the difference between a miscellaneous series and other types of series?

A miscellaneous series is different from other types of series such as arithmetic or geometric series because there is no pattern or formula to determine the terms. Other series have a constant difference or ratio between terms, making it easier to find the sum.

Can a miscellaneous series have negative terms?

Yes, a miscellaneous series can have negative terms. The terms can be any real numbers, positive, negative, or zero, as long as they do not follow a specific pattern.

Why is it important to find the sum of a miscellaneous series?

Finding the sum of a miscellaneous series is important because it allows us to determine the total value of the series. This can be useful in many real-life situations, such as calculating profits, expenses, or other types of data that do not follow a specific pattern.

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