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.
 A: Your expression is equal to $g(1/10)$, where 
$$g(x)=\frac{x}{2}\left((2)(1)+(3)(2)x+(4)(3)x^2+(5)(4)x^3+\cdots\right)$$
Take the power series $1+x+x^2+x^3+\cdots$ for $\frac{1}{1-x}$ and differentiate twice. We get $(2)(1)+(3)(2)x+(4)(3)x^2+\cdots$ if we do it term by term, and $\frac{2!}{(1-x)^3}$ if we do it the usual way. Thus 
$$g(x)=\frac{x}{2}\cdot\frac{2!}{(1-x)^3}$$
(when $|x|\lt 1$, and in particular at $x=1/10$). 
Remark: The idea generalizes. The $n$-th triangular nunber is $\binom{n}{2}$. The same idea can be used to calculate $\sum \binom{n}{k}x^n$ for $|x|\lt 1$ and fixed positive integer $k$.
A: $$S={1\over10}+{3\over100}+{6\over1000}+{10\over10000}+\cdots$$ $${S\over10}={1\over100}+{3\over1000}+{6\over10000}+\cdots$$ Subtracting, $${9S\over10}={1\over10}+{2\over100}+{3\over1000}+{4\over10000}+\cdots$$ Now do the same thing again, that is, divide by $10$ and subtract, to get $${81S\over100}={1\over10}+{1\over100}+{1\over1000}+\cdots={1\over9}$$
A: 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  $10/81$)
$= 1/9 [ 1 + 1/9 + 10/81 ]$ 
  $= 1/9 \times 100/81$
  $= 100/729$

