Sum to n terms of a series

[sum2nterms]

Hi,

I am using a self-study book, Differential Equations by K.A. Stroud and Dexter Booth. The book reviews sequences and series. Formulas were derived for the sum to n terms for the powers of natural numbers n, n^2, and n^3. The chapter problems for the most part have consisted of simplifying expressions to terms of n, n^2, or n^3 in order that the given formulas could be applied. However, I cannot get one of the problems properly simplified. The problem is given in this form:

Find the sum to n terms of

1/(1*2*3) + 3/(2*3*4) + 5/(3*4(5) ...

Obviously, this can be expressed as sum(1 to n) of (2n -1)/(n*(n+1)*(n+2). But I cannot reduce it appropriately in order to find the formula for n terms of the series. I have tried straightforward simplification as well as partial fractions.

Any ideas how to approach this?

Thanks!

 

[Deleted member 4993]

\(\displaystyle \frac{2n-1}{n(n+1)(n+2)} \, = \, - \, \frac{1}{2n} \, + \, \frac{3}{n+1} \, - \, \frac{5}{2(n+2)}\)

 

[sum2nterms]

I may not have stated the problem properly.

The formula for the sum of the natural numbers, 1 + 2 + 3 + 4 ... + n sum(r = 1 to n) of r where n is a natural number = n(n+1)/2.

So, for the sum of natural numbers 1 + 2 + 3 + 4 + 5, 5(5+1)/2 = 5*6/2 = 15. This is the case where n = 5.

I am looking for a similar formula for sum( 1 to n) of (2n-1)/(n(n+1)(n+2))

Also, it does not seem that (2n+1)/(n)(n+1)(n+2) = 1/2n + 3/(n+1) + 5/(2(n+2)).

 

[Deleted member 4993]

I may not have stated the problem properly.

The formula for the sum of the natural numbers, 1 + 2 + 3 + 4 ... + n sum(r = 1 to n) of r where n is a natural number = n(n+1)/2.

So, for the sum of natural numbers 1 + 2 + 3 + 4 + 5, 5(5+1)/2 = 5*6/2 = 15. This is the case where n = 5.

I am looking for a similar formula for sum( 1 to n) of (2n-1)/(n(n+1)(n+2))

Also, it does not seem that (2n+1)/(n)(n+1)(n+2) = - 1/2n + 3/(n+1) - 5/(2(n+2)).----- why seem?? Did you take pencil and paper and work it out?

 

[sum2nterms]

I use the word "seem" simply to be polite.

Yes, I tried two approaches to show the equality you posted, no joy.

 

[Denis]

Yes, I tried two approaches to show the equality you posted, no joy.

Why don't you simply substitute a value for n: you'll quickly see the equality.

> 1/(1*2*3) + 3/(2*3*4) + 5/(3*4*5) + 7/(4*5*6) + 9/(5*6*7) + 11/(6*7*8) .....

Where does this series come from anyway? You made it up? (I added 3 terms to yours)

Here's what that series looks like after doing the math on each term:
1/6 + 3/24 + 5/60 + 7/120 + 9/210 + 11/336 ....... 19/1320 .....

So you're trying to come up with a formula that gives the sum on the 1st n terms;
ok: what do you need to do to add the 1st 3 terms manually? Get my point?

Go have a look at sequence number A007531 here: www.research.att.com/~njas/sequences/index.html

 

[Deleted member 4993]

\(\displaystyle \frac{2n-1}{n(n+1)(n+2)} \, = \, - \, \frac{1}{2n} \, + \, \frac{3}{n+1} \, - \, \frac{5}{2(n+2)}\)

\(\displaystyle \, - \,\frac{1}{2n} \, + \, \frac{3}{n+1} \, - \, \frac{5}{2(n+2)}\)

\(\displaystyle = \, \frac{-(n+2)(n+1) + 6n(n+2) - 5n(n+1)}{2n(n+1)(n+2)}\)

\(\displaystyle = \, \frac{-(n^2+3n+2) + (6n^2+12n) - (5n^2+5n)}{2n(n+1)(n+2)}\)

Does it still seem to you the same way ...

By the way, what are the approaches that you did take??

 

[soroban]

Hello, sum2nterms!

\(\displaystyle \text{Find the sum to }n\text{ terms: }\;S \;=\;\frac{1}{1\cdot2\cdot3} + \frac{3}{2\cdot3\cdot4} +\frac{5}{3\cdot4\cdot5} + \hdots\)

\(\displaystyle \text{Obviously, this can be expressed as: }\:\sum^n_{k=1} \frac{2k -1}{k(k+1)(k+2)}\)

But I cannot reduce it appropriately in order to find the formula for n terms of the series.
I have tried straightforward simplification as well as partial fractions.

Your work on Partial Fractions must be faulty,
. . otherwise, you would have arrived at Subhotosh's result.
(It also worries me that you couldn't verify that he is correct.)

And do you know why we try Partial Fractions at all?
We hope to find that many/most terms will cancel out.
It seems that many teachers/textbooks don't explain this clearly.


\(\displaystyle \text{Since }\,\frac{1}{n(n+1)(n+2)} \;=\;\frac{\text{-}\frac{1}{2}}{n} + \frac{3}{n+1} + \frac{\text{-}\frac{5}{2}}{n+2},\;\text{ the series can be written like this:}\)


\(\displaystyle S \;=\;\left(\frac{\text{-}\frac{1}{2}}{1} + \frac{3}{2} - \frac{\frac{5}{2}}{3}\right) + \left(\frac{\text{-}\frac{1}{2}}{2} + \frac{3}{3} - \frac{\frac{5}{2}}{4}\right) + \left(\frac{\text{-}\frac{1}{2}}{3} + \frac{3}{4} - \frac{\frac{5}{2}}{5}\right) + \left(\frac{\text{-}\frac{1}{2}}{4} + \frac{3}{5} - \frac{\frac{5}{2}}{6}\right) + \hdots\)

. . \(\displaystyle \hdots +\left(\frac{\text{-}\frac{1}{2}}{n-2} + \frac{3}{n-1} - \frac{\frac{5}{2}}{n}\right) + \left(\frac{\text{-}\frac{1}{2}}{n-1} + \frac{3}{n} - \frac{\frac{5}{2}}{n+1}\right) + \left(\frac{\text{-}\frac{1}{2}}{n} + \frac{3}{n+1} - \frac{\frac{5}{2}}{n+2}\right)\)


Now examine the fractions by their denominators . . .

. . \(\displaystyle \text{There is one fraction with denominator 1: }\:\frac{\text{-}\frac{1}{2}}{1} \:=\:\text{-}\frac{1}{2}\)

. . \(\displaystyle \text{There are two fractions with denominator 2: }\:\frac{3}{2} - \frac{\frac{1}{2}}{2} \:=\:\frac{5}{4}\)

. . \(\displaystyle \text{There are three fractions with denominator 3: }\:\frac{\text{-}\frac{5}{2}}{3} + \frac{3}{3} - \frac{\frac{1}{2}}{3} \:=\:0\)

. . \(\displaystyle \text{There are three fractions with denominator 4: }\:\frac{\text{-}\frac{5}{2}}{4} + \frac{3}{4} - \frac{\frac{1}{2}}{4} \:=\:0\)
. . . . . . . . . \(\displaystyle \vdots\)
. . \(\displaystyle \text{There are three fractions with denominator }n\!:\:\frac{\text{-}\frac{5}{2}}{n} + \frac{3}{n} - \frac{\frac{1}{2}}{n} \:=\:0\)

. . \(\displaystyle \text{There are two fractions with denominator }n+1\!:\;\frac{\text{-}\frac{5}{2}}{n+1} + \frac{3}{n+1} \;=\; \frac{\frac{1}{2}}{n+1}\)

. . \(\displaystyle \text{There is one fraction with denominator }n+2\!:\;\frac{\text{-}\frac{5}{2}}{n+2}\)


\(\displaystyle \text{The remaining terms are: }\:S \;=\;\text{-}\frac{1}{2} + \frac{5}{4} + \frac{\frac{1}{2}}{n+1} - \frac{\frac{5}{2}}{n+2}\)

. . \(\displaystyle \text{which simplifies to: }\;\boxed{S \;=\;\frac{n(3n+1)}{4(n+1)(n+2)}} \quad\hdots \;There!\)

 

[sum2nterms]

Denis: As stated in the original post, the problem came from Differential Equations by K.A. Stroud and Dexter Booth.

Denis, Soroban, Subhotoah ... Thank you all for the rich responses. I do this in my spare time and unfortunately do not have as much time as I would like for it. I have struggled with this problem for some time, but I can see it is partly due to simple math mistakes. I did try the partial fractions approach and apparently did not do that correctly. I tried simplifying the right side of Subhotoah's response and did not get the left side. I then used the entire equation, treating both side the same, i.e. adding the same term to both sides, multiplying both side by the same term, etc. Obviously, I need to slow down a bit. I will have to take a look at the full solution and the partial ones, too. I have used partial fractions in the past in order to obtain integrable terms from an expression that is difficult to integrate. I am here to learn and I appreciate your time! Again, Thank You!