study-the-convergence-of-n-0-sin-pi-4n-2-1-

Question Number 52670 by maxmathsup by imad last updated on 11/Jan/19

s t u d y t h e c o n v e r g e n c e o f \sum_{n = 0}^{\infty} s i n (π \sqrt{4 n^{2} + 1})

Commented by maxmathsup by imad last updated on 12/Jan/19

w e h a v e π \sqrt{4 n^{2} + 1} = π \sqrt{4 n^{2} (1 + \frac{1}{4 n^{2}}}) = 2 n π \sqrt{1 + \frac{1}{4 n^{2}}} b u t w e h a v e

\sqrt{1 + x} = {(1 + x)}^{\frac{1}{2}} = 1 + \frac{x}{2} + \frac{\frac{1}{2} (\frac{1}{2} - 1)}{2} x^{2} + o (x^{3}) (x \to 0) \Rightarrow

= 1 + \frac{x}{2} - \frac{1}{8} x^{2} + o (x^{3}) \Rightarrow \sqrt{1 + \frac{1}{4 n^{2}}} = 1 + \frac{1}{8 n^{2}} - \frac{1}{8.16 n^{4}} + o (\frac{1}{n^{6}}) \Rightarrow

2 n π \sqrt{1 + \frac{1}{4 n^{2}}} = 2 n π + \frac{π}{4 n} - \frac{π}{32 n^{3}} + o (\frac{1}{n^{5}}) \Rightarrow s i n (π \sqrt{4 n^{2} + 1}) \sim s i n (\frac{π}{4 n}) \sim \frac{π}{4 n}

b u t Σ \frac{π}{4 n} d i v e r g e s \Rightarrow Σ s i n (π \sqrt{4 n^{2} + 1}) d i v e r g e s .

Answered by tanmay.chaudhury50@gmail.com last updated on 11/Jan/19

\frac{4 n^{2} + 1}{2} ⩾ \sqrt{4 n^{2} \times 1}

4 n^{2} + 1 ⩾ 4 n

\sqrt{4 n^{2} + 1} ⩾ \sqrt{4 n}

s i n (π \sqrt{4 n^{2} + 1}) \approx s i n (π \times 2 \sqrt{n})

s i n (\sqrt{n} \times 2 π)

1) n o w w h e n n i s p e r f e c t s q u a r e

s i n (\sqrt{n} \times 2 π) = 0

2) \sqrt{n} = I + f I = i n t e g e r

f = f r a c t i o n a l p a r t

s i n {(I + f) 2 π}

= s i n {I \times 2 π + f \times 2 π}

= s i n (f \times 2 π)

1 ⩾ s i n (2 π f) ⩾ - 1

s o \underset{n = 0}{\sum^{\infty}} s i n (π \sqrt{4 n^{2} + 1}) d o e s n o t c o n v e r g e

\underset{n = 0}{\sum^{\infty}} s i n (π \sqrt{4 n^{2} + 1}) o s c i l a t e s b e t w e e n \pm \infty

i h a v e t r i e d t o s o l v e \dots p l s c h e c k i s t h e l o g i c

t r u e \dots