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How can one use the comparison tests on

$$\sum_{n=3}^{\infty} \frac{\ln n}{4+n}$$

I've tried comparing with $\displaystyle b_n = \frac{\ln n }{n}$ but since $b_n > a_n$ and $b_n$ diverges (Integral test) I cannot assume $a_n$ diverges too.

Any help is highly appreciated!

Thank you.

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    $\begingroup$ $\forall n \ge 3, \ln (n) \gt 1$. $\endgroup$ Commented Dec 8, 2017 at 12:00

3 Answers 3

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For $n>e$ we have $\ln(n)>1$. Thus for such $n$ $$\frac{\ln(n)}{4+n}>\frac{1}{4+n}$$ and $$\sum_{n=3}^\infty\frac{\ln(n)}{4+n}>\sum_{n=3}^\infty\frac{1}{4+n}=-\sum_{n=1}^{6}\frac{1}{n}+\sum_{n=1}^\infty\frac{1}{n}$$ Thus the series diverges by comparison test with the harmonic series.

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One standard trick, when the denominator is going the wrong way:

$$\frac{\ln n}{4+n} >\frac{\ln n}{4+4n} >\frac{1}{4}\frac{\ln n}{n}.$$

Then either use integral test as you suggest, or continue:

$$\frac{1}{4}\frac{\ln n}{n}>\frac{1}{4}\frac{1}{n}.$$

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With equivalents:

The general term $\;\dfrac{\log n}{4+n}\sim_\infty\dfrac{\log n}n$, and the latter is a divergent Bertrand's series, since $$\sum_n\frac1{n^\alpha\log^\beta n}\quad\text{converges if and only if}\begin{cases}\alpha>1,\quad\text{or}\\\alpha=1\enspace\text{and}\enspace\beta>1.\end{cases}$$

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