- #1

- 986

- 9

## Homework Statement

Express (log n)

^{log n}as a power of n, and use the result to show that ∑1/(log n)

^{log n}converges.

## Homework Equations

blah blah

## The Attempt at a Solution

*n*=

^{p(n)}*(log n)*

^{log n}----> log[

*n*] = log[

^{p(n)}*(log n)*]

^{log n}*----> p(n) (log n) = (log n) log(log n)*

----> p(n) = log(log n)

----> ∑1/(log n)

----> log[

----> p(n) (log n) = (log n) log(log n)

----> p(n) = log(log n)

----> ∑1/(log n)

----> p(n) = log(log n)

----> ∑1/(log n)

^{(log n)}= ∑ 1/n^{log(log n)}*n*=^{p(n)}*(log n)*^{log n}----> log[

*n*] = log[^{p(n)}*(log n)*]^{log n}----> p(n) (log n) = (log n) log(log n)

----> p(n) = log(log n)

----> ∑1/(log n)

^{(log n)}= ∑ 1/n^{log(log n)}.... Now what? I can't think of any comparison test or anything for u

_{n}= 1/n

^{log(log n)}