abs/conditional convergence

[ijd5000]

I'm having trouble proving that the series from n=2 to infinity of (-1)^n/ln(n^3) is conditionally convergent

 

[ijd5000]

try the alternating series test.

\(\displaystyle \frac{1}{ln((n+1)^3} < \frac{1}{ln(n^3)}\) so the sequence of terms decreases monotonically to zero as \(\displaystyle n\rightarrow\infty\)

yes the AST shows that it converges, which test would show that the absolute value diverges? the root and ratio tests both are inconclusive.

 

[pka]

I'm having trouble proving that the series from n=2 to infinity of (-1)^n/ln(n^3) is conditionally convergent

Let \(\displaystyle a_n=(\ln(n))^{-1}\). Now show that \(\displaystyle (a_n)\) is a decreasing sequence and converges to zero.
That shows that \(\displaystyle \sum\limits_{k = 2}^\infty {\frac{{{{( - 1)}^k}}}{{\ln (k)}}} \) converges.

Then show that \(\displaystyle \sum\limits_{k = 2}^\infty {\frac{{{{( 1)}}}}{{\ln (k)}}} \) diverges.

 

[ijd5000]

Let \(\displaystyle a_n=(\ln(n))^{-1}\). Now show that \(\displaystyle (a_n)\) is a decreasing sequence and converges to zero.
That shows that \(\displaystyle \sum\limits_{k = 2}^\infty {\frac{{{{( - 1)}^k}}}{{\ln (k)}}} \) converges.

Then show that \(\displaystyle \sum\limits_{k = 2}^\infty {\frac{{{{( 1)}}}}{{\ln (k)}}} \) diverges.

is 1/ln(n) comparable with the harmonic series?

 

[lookagain]

ln(n) < n, n>=1

so 1/ln(n) > 1/n \(\displaystyle \ \) not for \(\displaystyle \ n \ge 1 \ \), as you stated above, but for (integers) n > 1, because ln(1) is not defined.

but we know > > 1/n < < diverges, so > > 1/ln(n) < < must diverge

We know \(\displaystyle \ \displaystyle\sum_{n = 2}^{\infty} \ \dfrac{1}{n} \ \ \) diverges, so \(\displaystyle \ \displaystyle\sum_{n = 2}^{\infty} \ \dfrac{1}{ln(n)} \ \ \) must diverge.

 

[ijd5000]

ln(n) < n, n>=2

so 1/ln(n) > 1/n

but we know 1/n diverges, so 1/ln(n) must diverge

Thank you for the explanation.