SOLUTION: please help me solve this sum on summation of series: {{{1/(1*2*3) + 3/(2*3*4) + 5/(3*4*5) + 7/(4*5*6)}}} + ···

 
The numerators are the odd numbers, and the nth odd integer
is 2n-1, so that will go on top in the nth term formula.  The 
denominator of the nth term formula is the product of n and 
the next two integers, (n+1) and (n+2), so the nth term is
 
 

 + ··· =  

Do you understand partial fractions?  If not post again
asking how.  But I'll assume you already know how.

We break the summand into partial fractions:



Now suppose we sum this just to some large positive integer K,
and then take the limit as K approaches infinity:

lim   
K->oo

Make three sums:

lim   
K->oo

Put the constant multipliers in front of the summations:

lim   
K->oo

Make the denominators in the second and third summands all be single
letters, by setting them equal to another letter:

In the second summation, let  or  and
in the third summation, let  or 

lim   
K->oo

lim   
K->oo

Now no change will result if we replace M and P by n:

lim   
K->oo

In the first summation we write out the first two terms, and
write the summation from n=3 to n=K



In the second summation we write out the first term and the last
term and write the summation from n=3 to n=K



In the third summation we write out the last two
terms and write the summation from n=3 to n=K



Now we let 

So, now the sums become:







So,

lim   
K->oo

becomes:

lim   
K->oo

or:

lim   
K->oo

Write the  as  and  as 

lim   
K->oo

Ths terms in S all cancel out, and we have

lim   
K->oo

The last three terms approach 0 as K approaches infinity.

Therefore the answer is 

Edwin