Alternating Series: Solving Homework Equations

[Jbreezy]

Homework Statement

Ʃ (-1)^n [ n+ln(n) / n-ln(n)] from n = 2 to infinity.

Homework Equations

I looked at the limit first because I thought lnn was very slow function. n would go faster.

The Attempt at a Solution

limit n --> ∞ [ n+ln(n) / n-ln(n)] = 1 so it diverges.
Limit is not 0 so it violates the one of the conditions? OK ? OR wrong?
Thanks,

 

[Dick]

Homework Statement

Ʃ (-1)^n [ n+ln(n) / n-ln(n)] from n = 2 to infinity.

Homework Equations

I looked at the limit first because I thought lnn was very slow function. n would go faster.

The Attempt at a Solution

limit n --> ∞ [ n+ln(n) / n-ln(n)] = 1 so it diverges.
Limit is not 0 so it violates the one of the conditions? OK ? OR wrong?
Thanks,

Yes, it doesn't converge. If $|x_n|$ doesn't converge to 0, then $\Sigma x_n$ doesn't converge. That's true whether the series is alternating or not. You still have to prove its limit isn't zero. Just saying 'slow function' doesn't do the job.

 

[Ray Vickson]

Homework Statement

Ʃ (-1)^n [ n+ln(n) / n-ln(n)] from n = 2 to infinity.

Homework Equations

I looked at the limit first because I thought lnn was very slow function. n would go faster.

The Attempt at a Solution

limit n --> ∞ [ n+ln(n) / n-ln(n)] = 1 so it diverges.
Limit is not 0 so it violates the one of the conditions? OK ? OR wrong?
Thanks,

USE PARENTHESES! What you have written is
\lim_{n \to \infty} n + \frac{\ln(n)}{n} - \ln(n) = 1 \, \leftarrow \text{ false}
Perhaps you meant
\lim_{n \to \infty} \frac{n + \ln(n)}{n - \ln(n)} = 1,
which is true. What would be so hard about writing (ln(n)+n}/(ln(n)-n), or [ln(n)+n]/[ln(n)-n], if that is what you really meant?