Summation of series questions
By Hal 2001 (P3046) on Friday, November 24, 2000 - 11:27 pm :

Hi

begin{displaymath}sum\_{r=1}^{20} frac{1}{r(r+1)}end{displaymath}



Any ideas, much appreciated!
Hal
By Dan Goodman (Dfmg2) on Saturday, November 25, 2000 - 01:00 am :
begin{displaymath}frac{1}{r(r+1)} = frac{1}{r} - frac{1}{r+1}end{displaymath}


so
begin{displaymath}sum\_{r=1}^{20} frac{1}{r(r+1)} = sum\_{r=1}^{20} Bigl(frac{1}{r}-frac{1}{r+1}Bigr)end{displaymath}


But in this version, the terms begin to cancel each other out:

begin{eqnarray*} && Bigl(frac{1}{1}-frac{1}{2}Bigr) + Bigl(frac{1}{2}-fr... ...r) - frac{1}{21} &=& 1 - frac{1}{21} &=& frac{20}{21} end{eqnarray*}






This technique is called the "method of differences" (I think, it's been a while since I had to use the name), in general if you want to sum f(r) from r=1 to n, and you can write f(r)=g(r)-g(r+1) then the sum comes to f(1)-f(n+1). In your specific case f(r)=1/r(r+1) and we can write it in this form, but this won't always work unfortunately. However, it's a useful method.

By Hal 2001 (P3046) on Saturday, November 25, 2000 - 09:51 am :

Hi Dan,

Thanks for the crystal clear explanation!
I now understand how to do the problem, and similar variations.

Warmest Thanks!
Hal

By Susan Langley (Sml30) on Saturday, November 25, 2000 - 10:00 am :

Another name for this method is telescoping a sum or something like that...

By Hal 2001 (P3046) on Saturday, November 25, 2000 - 01:09 pm :

Telescoping, neato name!

By Hal 2001 (P3046) on Tuesday, November 28, 2000 - 12:39 pm :

Not sure how to do:

begin{displaymath}sum\_{r=3}^{20} (r+2)^3end{displaymath}


By Tim Martin (Tam31) on Tuesday, November 28, 2000 - 01:19 pm :

Try expanding (r+2)3 and then using the standard results:

begin{eqnarray*} sum\_{1}^{n} x &=& frac{n(n+1)}{2} sum\_{1}^{n} x^2 &=& ... ...{n(n+1)(2n+1)}{6} sum\_{1}^{n} x^3 &=& frac{n^2(n+1)^2}{4} end{eqnarray*}







You might also find it helpful to use the fact that
begin{displaymath}sum\_{m}^{n} f(x) = sum\_{1}^{n} f(x) - sum\_{1}^{m-1} f(x)end{displaymath}


By Michael Doré (Md285) on Tuesday, November 28, 2000 - 03:17 pm :

Or just note that your question is:

53 + 63 + ... + 223
= (13 + 23 + ... + 223 ) - (13 + 23 + 33 + 43 )
= (222 232 )/4 - (42 52 )/4
= (11x23)2 - (2x5)2
= (11x23 - 10)(11x23 + 10)
= 243 x 263

whatever that works out to be.

By Tim Martin (Tam31) on Tuesday, November 28, 2000 - 03:51 pm :

Of course

begin{displaymath}sum\_{r=3}^{20} (r+2)^3 = sum\_{r=5}^{22} r^3end{displaymath}


which is easier than my original method. It won't always work, though.
By Hal 2001 (P3046) on Tuesday, November 28, 2000 - 04:25 pm :

Thanks Tim, Michael.