Writing an infinite series as the sum of the series I have $$ y= \sum_{x=1}^\infty \frac{k}{10^{k^{2}}} =0.1002000030000004...$$
I want to be able to write this in terms of a geometric progression.
I've tried doing $ 10y = 1+\frac{2}{10^3}+\frac{3}{10^8}+...$, but I can't seem to get it in the form where I can sum an infinite series.  I can see the difference between the powers of the terms is like 3,5,7... but I'm not sure how I could use this?
 A: There's no "nice" closed form. However, you can notice the following:
$$\sum_{k = 1}^{\infty} \frac{k}{10^{k^2}} = \sum_{n = 1}^{\infty}  \sum_{k = n}^{\infty} \frac{1}{10^{k^2}}$$
Furthermore, $$\sum_{k = 1}^{\infty} \frac{1}{10^{k^2}} = \frac{1}{2} \left(\theta_3 \left(0, \frac{1}{10}\right) - 1\right)$$
From this, one can derive similar expressions for $\displaystyle \sum_{k = 2}^{\infty} \frac{1}{10^{k^2}}$ and so forth by subtracting $\frac{1}{10^{k^2}}$
A: 
I can see the difference between the powers of the terms is like 3,5,7...

It is not geometric precisely because of this; the terms of a geometric series have a constant ratio, e.g. the series $\dfrac12,\dfrac14,\dfrac18,\dfrac{1}{16}$ has the constant ratio of $\dfrac12$. But in your series, the ratio changes, i.e. the ratio between the first two numbers is 
$$
\frac{2}{10000}\frac{10}{1} = \frac{2}{1000}
$$
but the next ratio is
$$
\frac{3}{10^9}\frac{10^4}{2} = \frac{3}{2}10^{-5}
$$
A: you could make it a geometric series with your observation. e.g.,
observe what
$$-\frac{k}{10^{k^2}}+\frac{k-1}{10^{(k-1)^2}}=\frac{-k+(k-1)(10^{2k-1})}{10^{k^2}}$$
so your series is 
$$\sum_{k=1}^{\infty}\sum_{n=1}^{k} \frac{10^{2n-1}-1)n-10^{2n-1}}{10^{n^2}}$$
which has 2 geometric progressions in the numerator, but the denominator cannot be modified to form any sort of geometric progression
