# 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?

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A little thing here, but the bottom of your summation should say $k = 1$. –  Soke Mar 10 at 4:24

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 $\frac{1}{10^{k^2}}$

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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 $$\frac{3}{10^9}\frac{10^4}{2} = \frac{3}{2}10^{-5}$$

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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

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