TITLE: How to sum $\frac1{1\cdot 2\cdot 3\cdot 4} + \frac4{3\cdot 4\cdot 5\cdot 6} + \frac9{5\cdot 6\cdot 7\cdot 8} + \cdots$ quickly?
QUESTION [8 upvotes]: The Problem:
$$\frac1{1\cdot 2\cdot 3\cdot 4} + \frac4{3\cdot 4\cdot 5\cdot 6} + \frac9{5\cdot 6\cdot 7\cdot 8} + \frac{16}{7 \cdot 8 \cdot 9 \cdot 10} + \cdots$$
Any smarter way to solve this may be within a minute or two?

I am adding my solution,If you are a student please don't read-on I find this particular problem to some sort of interesting so just try it once :-)
I started off with trying to find the $T_n$ th term
$$T_n = \frac{n^2}{(2n-1)(2n)(2n+1)(2n+2)}$$
$$ = \frac1{24} \times \biggl( \frac1{2n-1} + \frac3{2n+1} - \frac4{2n+2} \biggr)$$ 
$$ = \frac1{24} \times \biggl[\biggl( \frac1{2n-1} + \frac1{2n+1}\biggr) + 4 \times \biggl(\frac1{2n+1} - \frac1{2n+2} \biggr) \biggr]$$   
After this this becomes really easy,
  $$ S_\infty =   \frac1{24} \times \biggl[( 1 - \frac13 + \frac13 -  \frac15 +  \frac15 + \cdots ) + 4 \times (\ln 2 - \frac12)\biggr]$$
$$ = \frac16 \log_e 2 - \frac1{24}$$
But as you can see this approach is a bit tedious and it took me some 30 minutes to reduce the $T_n$ to that form. Any other smart approaches?

REPLY [3 votes]: Again this is trivial employing telescopy. After partial fraction decomposition the 
summand has the form $\rm\ a_1(n-1) - a_1(n) + c\ (a_1(n)+a_0(n))\ $ where $\rm\ a_0(n),\ a_1(n)\ $ are the coefficients of the even, odd part of the  $\rm log\ x\ $ power series, for $\rm\ x = 2\:.\:$ The sum of the first part $\rm\ a_1(n-1) - a_1(n)\ $ telescopes to $\rm\:a(0)\:$ and the sum of second part combines all the odd and even terms into the complete power series for $\rm\: log\ 2\:.\ $  Notice how very simple this approach is. In particular it does not require any knowledge of asymptotics of harmonic series  - as in the approach in Moron's answer.  
Generally no ingenuity is required to find such telescopy. It can be done  mechanically as follows.  Let $\rm\:S\:$ be the shift $\rm\: S\ a(n)\: \to\: a(n+1)\:,\: $ and let $\rm\ P(S)\ $ be a polynomial in $\rm\:S\:$ whose coefficients $\rm\:c\:$ are constant w.r.t. $\rm\:n\:,\ $ i.e. $\rm\ S\ c = c\:.\: $ Then $\rm\ \sum\ P(S)\ a(n)\ =\ \sum\ (P(S)-P(1))\ a(n)\ +\ P(1)\: \sum\: a(n)\:.\ $ The first sum telescopes since the Factor Theorem $\rm\: \Rightarrow\ S-1 \:|\: P(S)-P(1)\:.\ $ Therefore the sum computation reduces to that of the simpler $\rm\ \sum\: a(n)\:.$