Sum of series with triangular numbers Can you please tell me the sum of the seires
$ \frac {1}{10} + \frac {3}{100} + \frac {6}{1000} + \frac {10}{10000} + \frac {15}{100000} + \cdots $ 
where the numerator is the series of triangular numbers?
Is there a simple way to find the sum?
Thank you.

I thought I might add another derivation (devised by me). This one is long and involves dissecting the sequence into its simplest terms.

$1/10 + 3/100 + 6/1000 + \ldots$
$= 1/10 + (1+2)/100 + (1+2+3)/1000 + \ldots$ (from the definition of
  triangular numbers.)
$= 1/10 + 1/100 + 2/100 + 1/1000 + 2/1000 + 3/1000 + \ldots$
(by grouping terms with similar numerator together)
  $= (1/10 + 1/100 + 1/1000 + \ldots) + (2/100 + 2/1000 + \ldots) + (3/1000 +
 \ldots) + \ldots$ $= 1/9 + 2/90 + 3/900 + \ldots$
($1/9$ is a common factor)
$= 1/9 [ 1 + 2/10 + 3/100 + \ldots]$ $= 1/9 [ 1 + 1/10 + 1/10 + 1/100
 + 1/100 + 1/100 + \ldots ]$
(after rearranging the terms)
$= 1/9 [ 1 + (1/10 + 1/100 + 1/100 + \ldots) + (1/10 + 1/100 + 1/100 +
 \ldots) + (1/100 + \ldots) + \ldots ] $ $= 1/9 [ 1 + 1/9 + (1/9 + 1/90
 + 1/900 + 1/900 + \ldots) ]$
(the terms between the parentheses represent a geometric series
  whose sum is  $10/81$)
$= 1/9 [ 1 + 1/9 + 10/81 ]$ 
  $= 1/9 \times 100/81$
  $= 100/729$

