# Summation Confusion

1. Sep 5, 2009

### PhillipKP

1. The problem statement, all variables and given/known data

This is kind of a question regarding summation.

All logs are to base 2.

Given

$$A=\sum_{n=2}^{\infty}(n\log^{2}(n))^{-1}$$

Why does the the Author get

$$\sum_{n=2}^{\infty}\frac{\log A}{An\log^{2}(n)}=\log A$$
?

2. Relevant equations

3. The attempt at a solution

But working it out, I get

$$\sum_{n=2}^{\infty}\frac{\log A}{\sum_{n=2}^{\infty}(n\log^{2}(n))^{-1}n\log^{2}(n)}=\sum_{n=2}^{\infty}\frac{\log A}{\sum_{n=2}^{\infty}1}$$

Since $$A\approx1.013$$

$$log(A)\approx0.019$$

Therefore

$$\sum_{n=2}^{\infty}\frac{0.019}{\infty}=0$$

What did I do wrong?

Thanks for any help in advance.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 5, 2009

### javierR

A is already a summed expression. Therefore, you can pull it out of the sum:
$$\frac{logA}{A} \Sigma \left( n \log^2(n) \right)^{-1}$$, which is (logA/A)*A=logA.

3. Sep 5, 2009

### D H

Staff Emeritus
Too much help there, javier. A hint to pull a constant factor out of the sum would have been enough.

4. Sep 5, 2009

### PhillipKP

That was perfect thanks :)

Last edited: Sep 5, 2009
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