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How to determine whether or not the series

$$\sum_{n=1}^\infty\frac{\ln n+(-1)^n\sqrt{n}}{n\cdot \sqrt{n}}$$

is convergent?

I couldnt do such problems. What is the way? How should I think?

I edited the parentheses.

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Please check to make sure that I wrote what you intended. You should also read this to learn about accepting answers and why you should. –  Brian M. Scott Dec 25 '12 at 19:04
    
I doubt you would be asked a question in which the $n^{1/2}$ so obviously cancel. Maybe parentheses went astray? –  André Nicolas Dec 25 '12 at 19:09
    
I changed $ln n$, coded as "ln n", to $\ln n$, coded as "\ln n". That is standard. –  Michael Hardy Dec 25 '12 at 19:23
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3 Answers 3

up vote 3 down vote accepted

There has been a clarification, which changes the answer. We are looking at $$\sum_{n=1}^\infty\frac{\ln n+(-1)^nn^{\frac{1}{2}}}{n\cdot n^{\frac{1}{2}}}.$$ This can be split as $$\sum_{n=1}^\infty\frac{\ln n}{n^{3/2}} +\sum_{n=1}^\infty \frac{(-1)^n}{n}$$

The second is an alternating series, which converges.

For the first, note that if $n$ is large enough, then $\log n \lt n^{1/4}$. It follows that if $n$ is large enough, $$0\lt \frac{\ln n}{n^{3/2}}\lt \frac{n^{1/4}}{n^{3/2}}=\frac{1}{n^{5/4}}.$$ So the sum converges, by comparison with the "$p$-series" $\sum_{1}^\infty \frac{1}{n^{5/4}}$.

Thus our original series converges.

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By expanding out the brackets you cans split it into two series: $$\sum_{n=1}^\infty\frac{\left(\ln n+(-1)^n\right)n^{1/2}}{n\cdot n^{1/2}} = \sum_{n=1}^\infty{\frac{\ln n}{n}}+{\sum_{n=1}^\infty{\frac{(-1)^n}{n}}}$$ The second series is the alternating Harmonic, widely known to be convergent. This can be verified by the Alternating series test. The first series however, when you apply Ratio test, you will find that: $$\frac{a_{n+1}}{a_n}=\frac{\ln{(n+1)}}{\ln{n}}\frac{n}{n+1}$$ Which has a limit of $1$ as $n \to \infty$, so the ratio test in inconclusive. However, we can compare it term by term with the Harmonic series: $$\frac{1}{n}<\frac{\ln n}{n} \forall n > 1$$ Hence: $$\sum_{n=1}^\infty{\frac{\ln n}{n}} > \sum_{n=1}^\infty{\frac{1}{n}}$$ And the right hand side is known to diverge (Harmonic series), hence the entire series must diverge.

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Note that $$\sum_{n=1}^\infty\frac{\ln n+(-1)^nn^{\frac{1}{2}}}{n\cdot n^{\frac{1}{2}}}= \sum_{n=1}^\infty\frac{\ln n }{ n^{\frac{3}{2}} } + \sum_{n=1}^{\infty}\frac{ (-1)^n }{n}\,. $$

Now, the first series on the RHS converges by the integral test and the second series converges by the alternating series test. So the whole series converges.

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