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I have the following series:

$$\sum_{n=1}^{\infty}\left[{\frac{\sin{\frac 1 n}+\cos{\left({n\pi}\right)}}{n}}\right]$$

My textbook asks to determine the simple and absolute convergence of the series. Finding the simple (sorry, not sure what's the English term for it) convergence is easy enough:

$$\sum_{n=1}^{\infty}\left[{\frac{\sin{\frac 1 n}+\cos{\left({n\pi}\right)}}{n}}\right]=\sum_{n=1}^{\infty}\left[{\frac{\sin{\frac 1 n}}{n}}\right]+\sum_{n=1}^{\infty}\left[{\frac{\cos(n\pi)}{n}}\right]$$

The first series is positive and is asymptotic to $\frac 1 {n^2}$. The second one converges for the Leibniz rule.

My textbook though asks to check for the absolute convergence and it does the following passages:

$$\left|{\frac{\sin{\frac 1 n}+\cos{\left({n\pi}\right))}}{n}}\right|=\left|{(-1)^n\frac{1+(-1)^n\sin{\frac 1 n}}{n}}\right|=\frac{1+(-1)^n\sin{\frac 1 n}}{n}\sim\frac 1 n$$

I'm completely clueless about these passages. I am aware that both $\sin{\frac 1 x}$ oscillate between $-1$ and $1$ but I'm not sure if the textbook is grouping $(-1)$ and what rule it is following to do so. Any hints?

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  • $\begingroup$ is that greatest integer or simple brackets $\endgroup$ Sep 4, 2018 at 15:08
  • $\begingroup$ Cesare, you agreed in the first part that $\sin \frac{1}{n} \to 0.$ Therefore the alternating term in $\frac{1+(-1)^n\sin \frac{1}{n}}{n}$ vanishes. $\endgroup$
    – user376343
    Sep 4, 2018 at 15:13
  • $\begingroup$ $sin(\frac{1}{n})$ approaches zero as n becomes larger $\endgroup$ Sep 4, 2018 at 15:19
  • $\begingroup$ BTW: A very nice proof. Could you write the author of this textbook, please? $\endgroup$ Sep 4, 2018 at 15:24
  • $\begingroup$ @PrzemysławScherwentke it's by an Italian author Marco Bramanti: Analisi matematica 1 $\endgroup$
    – Cesare
    Sep 5, 2018 at 7:59

2 Answers 2

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Because $\cos(n\pi)=(-1)^n$ and $1=(-1)^{n}(-1)^n$, we have $$ \left|{\frac{\sin{\frac 1 n}+(-1)^n}{n}}\right| =\left|{(-1)^n\frac{(-1)^n\sin{\frac 1 n}+1}{n}}\right|=\frac{1+(-1)^n\sin{\frac 1 n}}{n}\sim\frac1n+(-1)^n\frac1{n^2}\sim\frac 1 n $$ But the harmonic series is divergent.

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Another approach is via the inequality $|a-b|\ge |a|-|b|.$ This gives

$$\frac{|(-1)^n-\sin(1/n)|}{n} \ge \frac{|(-1)^n|-|\sin(1/n)|}{n} = \frac{1-|\sin(1/n)|}{n}.$$

Since $\sin(1/n)\to 0,$ the term on the right is greater than $\dfrac{1/2}{n}$ for large $n,$ proving the series of absolute values diverges.

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