Question 146982
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)}}} + ···
<pre><font size = 4 color="indigo"><b> 
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
 
{{{(2n-1)/(n(n+1)(n+2))}}} 

{{{1/(1*2*3) + 3/(2*3*4) + 5/(3*4*5) + 7/(4*5*6)}}} + ··· = {{{sum((2n-1)/(n(n+1)(n+2)),n=1, infinity)}}} 

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:

{{{sum(((-1/2)/n + 3/(n+1) +(-5/2)/(n+2)),n=1, infinity)}}}

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

lim   {{{(sum(((-1/2)/n + 3/(n+1) +(-5/2)/(n+2)),n=1, n=K))}}}
K->oo

Make three sums:

lim   {{{(sum(((-1/2)/n),n=1, n=K) + sum((3/(n+1)),n=1, n=K)+sum(((-5/2)/(n+2)),n=1, n=K))}}}
K->oo

Put the constant multipliers in front of the summations:

lim   {{{((-1/2)*sum((1/n),n=1, n=K) + 3*sum((1/(n+1)),n=1, n=K)+(-5/2)*sum((1/(n+2)),n=1, n=K))}}}
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 {{{n+1=M}}} or {{{n=M-1}}} and
in the third summation, let {{{n+2=P}}} or {{{n=P-2}}}

lim   {{{((-1/2)*sum((1/n),n=1, n=K) + 3*sum((1/(M-1+1)),M-1=1, M-1=K)+(-5/2)*sum((1/(P-2+2)),P-2=1, P-2=K))}}}
K->oo

lim   {{{((-1/2)*sum((1/n),n=1, n=K) + 3*sum((1/M),M=2, M=K+1)+(-5/2)*sum((1/P),P=3, P=K+2))}}}
K->oo

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

lim   {{{((-1/2)*sum((1/n),n=1, n=K) + 3*sum((1/n),n=2, n=K+1)+(-5/2)*sum((1/n),n=3, n=K+2))}}}
K->oo

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

{{{sum((1/n),n=1,n=K)=1/1+1/2+sum((1/n),n=3,n=K)=3/2+sum((1/n),n=3,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

{{{sum((1/n),n=2,n=K+1)=1/2+sum((1/n),n=3,n=K)+1/(K+1))}}}

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

{{{sum((1/n),n=3,n=K+2)=sum((1/n),n=3,n=K)+1/(K+1)+1/(K+2))}}}

Now we let {{{sum((1/n),n=3,n=K)=S}}}

So, now the sums become:

{{{sum((1/n),n=1,n=K)=3/2+S}}}

{{{sum((1/n),n=2,n=K+1)=1/2+S+1/(K+1)}}}

{{{sum((1/n),n=3,n=K+2)=S+1/(K+1)+1/(K+2))}}}

So,

lim   {{{((-1/2)*sum((1/n),n=1, n=K) + 3*sum((1/(n+1)),n=1, n=K)+(-5/2)*sum((1/(n+2)),n=1, n=K))}}}
K->oo

becomes:

lim   {{{((-1/2)*(3/2+S)+ 3*(1/2+S+1/(K+2))+(-5/2)*(S+1/(K+1)+1/(K+2)))}}}
K->oo

or:

lim   {{{(-3/4-S/2+3/2+3S+3/(K+2)-5S/2-5/(2(K+1))-5/(2(K+3)))}}}
K->oo

Write the {{{3S}}} as {{{6S/2}}} and {{{3/2}}} as {{{6/4}}}

lim   {{{(-3/4-S/2+6/4+6S/2+3/(K+2)-5S/2-5/(2(K+1))-5/(2(K+3)))}}}
K->oo

Ths terms in S all cancel out, and we have

lim   {{{(3/4+3/(K+2)-5/(2(K+1))-5/(2(K+3))))}}}
K->oo

The last three terms approach 0 as K approaches infinity.

Therefore the answer is {{{3/4}}}

Edwin</pre>