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Show $\displaystyle\sum_{n=1}^\infty \frac{\mathrm{ln}(n)}{n^x}$ is convergent for $x>1$.

I've tried ratio test and I get $$\underset{n \rightarrow \infty}{\mathrm{lim}} \left\lvert \frac{\mathrm{ln}(n+1)n^x}{(n+1)^x\mathrm{ln}(n)} \right\rvert = 1$$ so I can't really conclude anything there.

So then I look at the limit comparison test: $$\underset{n \rightarrow \infty}{\mathrm{lim}} \frac{\mathrm{ln}(n)}{n^x}=0$$ so this is not helpful.

The end result I am really trying to prove here is that the Riemann zeta function converges uniformly on $[a, +\infty)$ (which I have done already) where $a>1$ and that $\displaystyle\zeta'(x) = \sum_{n=1}^{\infty} \frac{- \mathrm{ln}(n)}{n^x}$ for $x>1$.

The way that I showed the Riemann zeta function converges uniformly on $[a,\infty)$ was by using the Weierstrass $M$-test and noting that $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^x}$ converges for $x>1$ by the $p$-test.

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I am the ignorant! thank you all for the help, will study what you guys have to say. –  Starlight Feb 17 '13 at 11:18

4 Answers 4

up vote 1 down vote accepted

Convergence of this series is provable using 'not helpful' limit $$\lim_{n\to\infty} \frac{\ln n}{n^x}=0$$ for all $x>0$.

Let $x>1$, and take $c$ such that $0<c<x-1$. By previous limit, there exists $N$ such that $$n>N \implies \left| \frac{\ln n}{n^{x-1-c}} \right|<1$$ . So if $n>N$ then $$\frac{\ln n}{n^x}<\frac{1}{n^{1+c}}.$$ So $$\sum_{n=1}^\infty \frac{\ln n}{n^x}<\sum_{n=1}^N \frac{\ln n}{n^x}+\sum_{n>N} \frac{1}{n^{1+c}}$$

Since each term of series (on previous inequality) is non-negative, so by comparison test, this series converges.

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Using the integral convergence test with $f(n)=\frac{\ln(n)}{n^{x}}$, we have:

$$\int_{1}^{\infty}\frac{\ln(n)}{n^{x}}\:dn=\lim_{a\to\infty}\int_{1}^{a}\frac{\ln(n)}{n^{x}}\:dn=\lim_{a\to\infty}\left[-\frac{1+(x-1)\ln(n)}{n^{x-1}(x-1)^2}\right]_{1}^{a}=\left(-\frac{1}{(x-1)^{2}}\right)-\left(0\right)$$

Therefore, for $x>1$, the integral converges and thus so does the series.

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Yet another way: Compare with $$\sum \frac{1}{n^a}$$ for some $1 < a < x$. (Limit comparison test is the easiest here.)

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The function $f(n) = \frac{\ln(n)}{n^x}$ is positive and becomes decreasing for $n > M$ ($M$ depends on $x$). Use the Cauchy condensation test:

$$ 2^n\frac{\ln(2^n)}{2^{xn}} = \frac{n\ln 2}{2^{(x-1)n}} $$

Since $x > 1$, we have $x - 1 > 0$, $2^{x-1} > 1$ and the series converges.

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