Test-for-convergence-of-the-series-sin-nx-n-2-n-1-

Question Number 6268 by sanusihammed last updated on 21/Jun/16

Test for convergence of the series ∞ Σ ((sin(nx))/n^2 ) n = 1

$${Test}\:{for}\:{convergence}\:{of}\:{the}\:{series}\: \\ $$$$\infty \\ $$$$\Sigma\:\:\:\:\:\frac{{sin}\left({nx}\right)}{{n}^{\mathrm{2}} } \\ $$$${n}\:=\:\mathrm{1}\: \\ $$

Commented by FilupSmith last updated on 21/Jun/16

sin(ax+b)∈(−1, 1) ∴ ∃k:n^k ≥sin(ax+b), k,n∈Z^+ lim_(n→∞) ((sin(nx))/n^2 ) = ((constant)/∞), constant∈(−1, 1) =0 ∴ limit converges

$$\mathrm{sin}\left({ax}+{b}\right)\in\left(−\mathrm{1},\:\mathrm{1}\right) \\ $$$$\therefore\:\exists{k}:{n}^{{k}} \geqslant\mathrm{sin}\left({ax}+{b}\right),\:\:\:{k},{n}\in\mathbb{Z}^{+} \\ $$$$ \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{sin}\left({nx}\right)}{{n}^{\mathrm{2}} }\:=\:\frac{\mathrm{constant}}{\infty},\:\:\:\mathrm{constant}\in\left(−\mathrm{1},\:\mathrm{1}\right) \\ $$$$=\mathrm{0} \\ $$$$ \\ $$$$\therefore\:\mathrm{limit}\:\mathrm{converges} \\ $$

Commented by sanusihammed last updated on 21/Jun/16

$${Thanks} \\ $$

Answered by Yozzii last updated on 21/Jun/16

Since ∣sin(nx)∣≤1 for all n,x∈R ⇒((∣sin(nx)∣)/n^2 )≤(1/n^2 ) ⇒Σ_(n=1) ^∞ ((∣sin(nx)∣)/n^2 )≤Σ_(n=1) ^∞ (1/n^2 ) Now, Σ_(n=1) ^∞ (1/n^2 ) is convergent. (ζ(2)=Σ_(n=1) ^∞ (1/n^2 )=(π^2 /6)). Since Σ_(n=1) ^∞ (1/n^2 ) is convergent and Σ_(n=1) ^∞ ((∣sin(nx)∣)/n^2 )≤Σ_(n=1) ^∞ (1/n^2 ), then the comparison test states that Σ_(n=1) ^∞ ((∣sin(nx)∣)/n^2 )=Σ_(n=1) ^∞ ∣((sin(nx))/n^2 )∣ is convergent. Since for an absolutely convergent series Σ∣a_n ∣ implies that Σa_n is convergent, then Σ_(n=1) ^∞ ((sin(nx))/n^2 ) is convergent.

$${Since}\:\mid{sin}\left({nx}\right)\mid\leqslant\mathrm{1}\:{for}\:{all}\:{n},{x}\in\mathbb{R} \\ $$$$\Rightarrow\frac{\mid{sin}\left({nx}\right)\mid}{{n}^{\mathrm{2}} }\leqslant\frac{\mathrm{1}}{{n}^{\mathrm{2}} } \\ $$$$\Rightarrow\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mid{sin}\left({nx}\right)\mid}{{n}^{\mathrm{2}} }\leqslant\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} } \\ $$$${Now},\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:{is}\:{convergent}.\:\left(\zeta\left(\mathrm{2}\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} }=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\right). \\ $$$${Since}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:\:{is}\:{convergent}\:{and}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mid{sin}\left({nx}\right)\mid}{{n}^{\mathrm{2}} }\leqslant\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} }, \\ $$$${then}\:{the}\:{comparison}\:{test}\:{states}\:{that}\: \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mid{sin}\left({nx}\right)\mid}{{n}^{\mathrm{2}} }=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\mid\frac{{sin}\left({nx}\right)}{{n}^{\mathrm{2}} }\mid\:{is}\:{convergent}.\:{Since} \\ $$$${for}\:{an}\:{absolutely}\:{convergent}\:{series}\:\Sigma\mid{a}_{{n}} \mid \\ $$$${implies}\:{that}\:\Sigma{a}_{{n}} \:{is}\:{convergent},\:{then} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{sin}\left({nx}\right)}{{n}^{\mathrm{2}} }\:{is}\:{convergent}. \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by sanusihammed last updated on 21/Jun/16

$${Thanks}\:{for}\:{help} \\ $$

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