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I have to find the radius of convergence of the power series $\sum a_n z^n$ where $ a_n =$ number of divisors of $n^{50}$.

Options available are:

  1. $1$

  2. $50$

  3. $\frac{1}{50}$

  4. $0$

Please suggest how to proceed.

Using the fact that $ d(n)\leq n$, we have $d(n^{50})\leq n^{50}$.

Using the First Comparison Test, the series on the right converges if $\mid{z}\mid< 1$ and however does not converge if $\mid{z}\mid\geq 1$ and hence so does the series on the left. Radius of Convergence is 1.

[Now for $a_n =n$, then $ R = \lim_{n\rightarrow\infty}\frac {a_n}{a_{n+1}} =1. $

So,the series on the right converges for $\mid z \mid<1$]

Now if $ z=1$, then the series $\sum a_n z^n$ takes the form $\sum d(n^{50})$ and since

$ d(n^{50})\geq 1 $, by evoking the comparison test again, the series diverges for $z=1$.

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Please tell us your thoughts on the problem. Where are you stuck? –  Antonio Vargas May 18 '12 at 15:16
    
@AntonioVargas: Please see my working above. I generally write out my working but here as you can see ...I am totally stuck. –  preeti May 18 '12 at 16:33
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2 Answers

up vote 3 down vote accepted

Hint: Use the standard inequality $d(n) \leq n$ to compare the coefficients of this series with another one. Obtain a lower bound on the radius of convergence.

Hint 2: Once you have found a lower bound on the radius of convergence, test the convergence at a point on the boundary to show that the bound you found above actually is the exact radius of convergence.

(Here $d(n)$ is the number of divisors of $n$.)

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@preeti, yes, $d(n)$ is the number of divisors of $n$. Let me know how far you can get with the hint, and if you need another. –  Antonio Vargas May 18 '12 at 16:47
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Please see if it is correct. –  preeti May 18 '12 at 17:07
1  
@preeti yes, you are almost there. But, the comparison test does not give you the exact radius of convergence, only a lower bound. For example, $1/n! \leq 1$ for all $n$, but $\sum z^n$ converges only for $|z| < 1$ and $\sum z^n/n!$ converges everywhere. So, if $R$ is the radius of convergence of your series, you have shown that $R \geq 1$. To show $R = 1$, you now have to show that the series diverges somewhere on the circle $|z| = 1$. –  Antonio Vargas May 18 '12 at 17:12
1  
@preeti, you have to show that the series $\sum a_n z^n$ diverges for $z=1$. What do you mean by "the series on the right hand side takes the form $1+1+1+\cdots$"? –  Antonio Vargas May 18 '12 at 19:41
1  
@preeti, let us continue this discussion in chat –  Antonio Vargas May 18 '12 at 20:08
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Ok, so from what you did it seems clear that as $\,n\to\infty\,$, the number of divisors of $\,n\,$ , and thus also of $\,n^{50}\,$ , is not bounded, so...?

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