Math Help: Need to find convergence of this series.

[rob3rt]

does anyone know how to figure out what does this seriesconverges to....

2(1) + 3(1+2) + 4(1+2+3) + 5(1+2+3+4) + .. + (n-1)[1+2+3+...+(n-2)] = ?

for example,
1+2+3+4+5+....+n = n(n+1)/2

 

[ApacheXMD]

i forgot all the stuff I learned in Diff. Eq.


-patchy

 

[MadMark]

You can try to do this:

Re-group the expression to the follow:

2(1) + 3(1) + 4(1) + ..................... + (n-1)(1)
3(2) + 4(2) + ......................+ (n-1)(2)
.
.
.
(n-2)(n-3) + (n-1)(n-3)
+ (n-1)(n-2)
------------------------------------------------------------------
something I am too lazy to add up now. but you get the point. it uses the example formula you provided though.

 

[MadMark]

sorry for the mess but I don't know if there's a way to line the stuffs up.

 

[JayHu]

2(1) + 3(1+2) + ...
is not summation of (n-1)(1+2+3+..+n-2) it is summation (n+1)(1+2+...+n)
so..
summation from n=1 to infinity of (n+1)(1+2+...+n) = summation n=1 to infinity of (n/2)(n+1)^2 .. which wont converge.
since the limit of the function is not 0, then the series can't converge.
mmm... calculus II...