1. ## Convergence and Divergence

Show that the series ((ln n)^q)/n^p from n = 1 to infinity converges if p>1 where q is any real number.

I said that
((ln n)^q)/n^p = ((ln n)^q)/n X 1/(n^(p-1)) < 1/n^(p-1)

From the knowledge of p-series, the original series will converge if p>2. Shouldn't I instead get that p>1?

Also, is there a way to do this question using the limit comparison test? I attempted to use a = ((ln n)^q)/n^p and b = 1/n^p. I got the limit as
(ln n)^q which obviously diverges as n approaches infinity. Where did I make a mistake in this approach?

Thanks.

2. Originally Posted by dalbir4444
Show that the series ((ln n)^q)/n^p from n = 1 to infinity converges if p>1 where q is any real number.

I said that
((ln n)^q)/n^p = ((ln n)^q)/n X 1/(n^(p-1)) < 1/n^(p-1)

From the knowledge of p-series, the original series will converge if p>2. Shouldn't I instead get that p>1?

Also, is there a way to do this question using the limit comparison test? I attempted to use a = ((ln n)^q)/n^p and b = 1/n^p. I got the limit as
(ln n)^q which obviously diverges as n approaches infinity. Where did I make a mistake in this approach?
You need to use the fact that "log n tends to infinity slower than any positive power of n".

Given p>1, let $\displaystyle \delta = \tfrac12(p-1)$. Then, for all sufficiently large n, $\displaystyle \ln n <n^{\delta/q}$. It follows that $\displaystyle \frac{(\ln n)^q}{n^p} < \frac{n^\delta}{n^p} = \frac1{n^{1+\delta}}$, and you can then use the comparison test.

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### q) Lnn

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