# 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.

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Fixed $\LaTeX$ and problem statement. –  Ahaan S. Rungta Jul 16 at 17:39
@AhaanRungta you do not need to comment that here. –  picakhu Jul 16 at 21:44

$$S={1\over10}+{3\over100}+{6\over1000}+{10\over10000}+\cdots$$ $${S\over10}={1\over100}+{3\over1000}+{6\over10000}+\cdots$$ 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 $${81S\over100}={1\over10}+{1\over100}+{1\over1000}+\cdots={1\over9}$$

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Thanks for the fast response. Your manipulation is quite nice and makes the question too easy to solve. –  Mustafa Saad Jul 16 at 13:46
Indeed, WolframAlpha agrees –  Ryan Jul 16 at 15:50

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 $k$.

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Good answer, but one mistake. Differentiating $\frac1{1-x}$ twice gives you $\frac2{(1-x)^3}$ or $g(x)=\frac x{(1-x)^3}$. Then plugging in $\frac1{10}$ matches Gerry's answer. –  Mike Jul 16 at 13:49
Thanks, fixed! I tend to forget about constants. –  André Nicolas Jul 16 at 13:58
It was more than just the constant. I think you only differentiated once. The exponent is still wrong. –  Mike Jul 16 at 19:57
Thank you, exponent also fixed. Bad day! –  André Nicolas Jul 16 at 20:16

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 $10/81$)

$= 1/9 [ 1 + 1/9 + 10/81 ]$ $= 1/9 \times 100/81$ $= 100/729$

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