Using Leibniz on \sum_{n=1}^\infty \sin(\pi \sqrt{n^2+1})

airesex2

airesex2

Answered question

2022-04-23

Using Leibniz on n=1sin(πn2+1)

Answer & Explanation

utloverej

utloverej

Beginner2022-04-24Added 15 answers

Let n2+1=n+d(n). Then
sin(πn2+1)=sin(π(n+d(n))
=sin(πn)cos(πd(n))+cos(πn)sin(πd(n))
=(1)nsin(πd(n))
Also
d(n)=n2+1n
=(n2+1n)n2+1+nn2+1+n
=1n2+1+n
so that 0<d(n)<12n and d(n) is decreasing so that, since sinx is increasing for π2<x<π2,  sin(πd(n)) is also decreasing.
Therefore n=0(1)nsin(πd(n)) converges by the alternating series test.
eldgamliru9x

eldgamliru9x

Beginner2022-04-25Added 16 answers

sin(πn2+1)=sin(πn1+1n2)sin(πn(1+12n2))=(1)nsin(π2n)
(1)nπ2n

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?