## Loser66 one year ago Old problem confirm $\sum_{n=1}^\infty \dfrac{z^{n^2}}{n}=\sum_{k=1}^\infty a_kz^k$ Then $$a_k =??$$ Please, help

1. Loser66

@SithsAndGiggles

2. anonymous

If you're replacing $$n^2$$ with $$k$$, that means you have this pattern: $\begin{array}{c|cccc} n&1&2&3&4&5&\cdots\\ \hline k&1&4&9&16&25&\cdots \end{array}$ So while $$a_n$$ is defined for all positive integers $$n$$, $$a_k$$ is only defined for perfect squares. In other words, you can transform $$a_n$$ to $a_k=\begin{cases}a_{\sqrt n}&\text{ for }n\in\{1,4,9,16,25,\ldots\}\\0&\text{otherwise}\end{cases}$

3. Loser66

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4. Loser66

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5. Loser66

Since I have to find$$lim sup \sqrt[k] a_k$$, hence a_k must be clear.

6. anonymous

Yes, our definitions of $$a_k$$ are identical.

7. anonymous

Notice $a_n=\frac{1}{n}~~\implies~~a_{\sqrt n}=\frac{1}{\sqrt n}$ If $$n$$ is a perfect square, i.e. $$n^2=k$$, then $$a_k=a_{\sqrt n}=\dfrac{1}{\sqrt{n}}=\dfrac{1}{k}$$.

8. anonymous

Sorry, swap $$n$$ and $$k$$ in the last set of equations :)

9. Loser66

I interpret it like this $$\sum_{n=1}^\infty \dfrac{z^{n^2}}{n}= z + \dfrac{z^4}{2}+\dfrac{z^9}{3}+...$$ compare to $$\sum_{k=1}^\infty a_kz^k = a_1z +a_2z^2 +a_3z^3 +a_4z^4+....+a_9z^9+...$$ hence the equivalent part is a_1 =1 a_4 = 1/2 a_9 = 1/3 all other a_j =0

10. Loser66

a_1 is k =1 when n=1 n^2 =1 a_4 = 1/2 when n =2, n^2 =4, hence k = 1/ sqrt n

11. Loser66

But I am not sure about the terminology I should put there.

12. Loser66

if $$a_k = \dfrac{1}{\sqrt k}$$ , then $$limsup\sqrt[k]a_k= limsup\dfrac{1}{\sqrt[2k]k}$$ and by squeeze theorem it is =1 still but if $$a_k = 1/k,$$ we still have limsup =1 that is why I need confirm.

13. imqwerty

ok so i did this - $\sum_{n=1}^{\infty}\frac{ z^\left( n^2 \right) }{ n }=\sum_{k=1}^{\infty}a _{k}z^k$ $d \frac{ \sum_{n=1}^{\infty} z^\left( n^2 \right)n^\left( -1 \right)}{ dz}=d \frac{ \sum_{k=1}^{\infty}a _{k}z^k }{ dz}$ $\sum_{n=1}^{\infty}\frac{ n^2z^\left( (n+1)(n-1) \right) }{ n }=\sum_{k=1}^{\infty}a _{k}\times k \times z^{k-1}$ $\sum_{n=1}^{\infty}z ^{n+1}=\sum_{k=1}^{\infty}a_{k}$ ?

14. Loser66

And you know that when we turn it to k, everything must be calculated w.r.t k, not n anymore. n is there to show the relationship between the original one to the new one

15. Loser66

@imqwerty what are you doing?

16. imqwerty

i differentiated the summation on both sides of the equation

17. Loser66

why do we have to do that?

18. imqwerty

i did that so as to remove/factor out the n^2 from the z

19. Loser66

No need, but I think I got it, just turn the original one like this $\dfrac{z^{n^2}}{\sqrt{n^2}}$ hence to k, it will be $$\dfrac{z^k}{\sqrt{k}}$$

20. Loser66

Am I right?

21. Loser66

Hence $$a_k = 1/\sqrt k$$ if k = n^2

22. imqwerty

yes :)

23. Loser66

Thank you so much, both you @SithsAndGiggles and @imqwerty . I do appreciate:)

24. imqwerty

np :) i like ur questions but can we get the same result using my method? :/

25. Loser66

I don't think so, since a_k is a coefficient but in your method a_k is a series of variable z.

26. imqwerty

ok thanks :)

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