Question

For which p∈R is the infinite series
∑k=2∞=k(−1)k(k−1)2logp(k)
convergent? conditionally convergent? absolutely convergent?

Answer 1

mind drawing it out?

Answer 2

Calculate the ratio between the \(n\)th term and the \((n+1)\)th term.

Answer 3

If \(\left|\dfrac{n+1}n\right|>1\), divergent. If it's <1, convergent.

Answer 4

if it's =1, conditionally convergent.

Answer 5

sorry.. it should have been

\[\sum_{k=2}^{\infty} =\frac{ k(-1)^{k} }{ (k-1)^{2}\log ^{p} (k)}\]

so, kc_kennylau, your comment is still valid ?

Answer 6

um...

Answer 7

There are many convergence tests, I only know one of them...

Answer 8

The answer depends on p.

Answer 9

For this one use the integral test and the problem that I solved for you before.

Answer 10

no its an alternating series, so use alternating series test

Answer 11

The question is not only about convergence only. Please read the question.

Answer 12

Try to prove that if p>1, then the series is absolutely convergent.
If \( p \le 1\) one can show that is not absolutely convergent but convergent. Hence it is conditionally convergent.

Answer 13

If the ratio is 1 then it is inconclusive, not convergent.

Answer 14

Look at the series:
\[\sum_n \frac{1}{n}\]
The harmonic series is divergent but the ratio test gives 1.

Answer 15

Also, that is sloppy notation. You are not taking:
\[{\lim_{n \rightarrow \infty} \left| \frac{n+1}{n}\right|}\]
If your sum is:
\[\sum_n a_n\]
You're considering:
\[\lim_{n \rightarrow \infty} \left| \frac{a_{n+1}}{a_n}\right|<1\]
which tells your series is decreasing and therefore convergent.

Answer 16

So evaluate that limit for your series. From there impose the restriction and you'll find your condition on p. That answers your question.