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

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

T e s t f o r c o n v e r g e n c e o f t h e s e r i e s

\infty

Σ \frac{s i n (n x)}{n^{2}}

n = 1

Commented by FilupSmith last updated on 21/Jun/16

\sin (a x + b) \in (- 1, 1)

∴ \exists k : n^{k} ⩾ \sin (a x + b), k, n \in Z^{+}

\lim_{n \to \infty} \frac{\sin (n x)}{n^{2}} = \frac{constant}{\infty}, constant \in (- 1, 1)

= 0

∴ limit converges

Commented by sanusihammed last updated on 21/Jun/16

T h a n k s

Answered by Yozzii last updated on 21/Jun/16

S i n c e ∣ s i n (n x) ∣⩽ 1 f o r a l l n, x \in R

\Rightarrow \frac{∣ s i n (n x) ∣}{n^{2}} ⩽ \frac{1}{n^{2}}

\Rightarrow \underset{n = 1}{\sum^{\infty}} \frac{∣ s i n (n x) ∣}{n^{2}} ⩽ \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2}}

N o w, \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2}} i s c o n v e r g e n t . (ζ (2) = \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2}} = \frac{π^{2}}{6}) .

S i n c e \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2}} i s c o n v e r g e n t a n d \underset{n = 1}{\sum^{\infty}} \frac{∣ s i n (n x) ∣}{n^{2}} ⩽ \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2}},

t h e n t h e c o m p a r i s o n t e s t s t a t e s t h a t

\underset{n = 1}{\sum^{\infty}} \frac{∣ s i n (n x) ∣}{n^{2}} = \underset{n = 1}{\sum^{\infty}} ∣ \frac{s i n (n x)}{n^{2}} ∣ i s c o n v e r g e n t . S i n c e

f o r a n a b s o l u t e l y c o n v e r g e n t s e r i e s Σ ∣ a_{n} ∣

i m p l i e s t h a t Σ a_{n} i s c o n v e r g e n t, t h e n

\underset{n = 1}{\sum^{\infty}} \frac{s i n (n x)}{n^{2}} i s c o n v e r g e n t .

Commented by sanusihammed last updated on 21/Jun/16

T h a n k s f o r h e l p