Solve Summation Confusion: Homework Equations & Attempted Solution

[PhillipKP]

Homework Statement

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

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

Homework Equations

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.

 

[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.

 

[D H]

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

 

[PhillipKP]

That was perfect thanks :)