Question

prove one of this identities

Answer 1

\[\huge \color{red}{\sum_{n=1}^{\infty}3^{n-1}\sin ^3(\frac{ a }{ 3^n })=\frac{ 1 }{ 4 }( a-\sin a) }\]

Answer 2

\[\huge \color{blue}{\sum _{k=1}^ \infty \tan ^{-1}\frac{ 1 }{ 2k^2 }=\tan ^{-1}\frac{ n }{ n+1 }}\]

Answer 3

This section is about telescopic sums and products in trigonometry

Answer 4

so i am excpeting a lot of terms in between to cancel

Answer 5

I don't really see anything cancelling out in the first series so it's probably not telescopic.

Answer 6

important identity for blue
\[\Large \tan^{-1}u-\tan^{-1}v=\tan^{-1}\frac{ u-v }{ 1+uv }\]

Answer 7

\[\Large 4\sin^3x=\sin x-\sin 3x\]

Answer 8

we are almost there

Answer 9

\[
\sum_{n=1}^{\infty}3^{n-1}\sin^{3}\left(\frac{a}{3^{n}}\right)
\]


\[
\sum_{n=1}^{\infty}3^{n-1}\frac{1}{4}\left(3\sin\frac{a}{3^{n}}-\sin\frac{a}{3^{n-1}}\right)
\]


few terms:

\[
\left(\frac{3}{4}\sin\frac{a}{3}-\frac{1}{4}\sin a\right)+3\left(\frac{3}{4}\sin\frac{a}{9}-\frac{1}{4}\sin\frac{a}{3}\right)+9\left(\frac{3}{4}\sin\frac{a}{27}-\frac{1}{4}\sin\frac{a}{9}\right)+\cdots + \\ 3^{n-2}\left(\frac{3}{4}\sin\frac{a}{3^{n-1}}-\frac{1}{4}\sin\frac{a}{3^{n-2}}\right)+3^{n-1}\left(\frac{3}{4}\sin\frac{a}{3^{n}}-\frac{1}{4}\sin\frac{a}{3^{n-1}}\right)
\]

\[
\left(\cancel{\frac{3}{4}\sin\frac{a}{3}}-\frac{1}{4}\sin a\right)+\cancel{3\left(\frac{3}{4}\sin\frac{a}{9}-\frac{1}{4}\sin\frac{a}{3}\right)+9\left(\frac{3}{4}\sin\frac{a}{27}-\frac{1}{4}\sin\frac{a}{9}\right)}+\cdots + \\ \cancel{3^{n-2}\left(\frac{3}{4}\sin\frac{a}{3^{n-1}}-\frac{1}{4}\sin\frac{a}{3^{n-2}}\right)}+3^{n-1}\left(\frac{3}{4}\sin\frac{a}{3^{n}}-\cancel{\frac{1}{4}\sin\frac{a}{3^{n-1}}}\right) =
\]
\[-\frac{1}{4}\sin a + \frac{3^n}{4}\sin \frac{a}{3^n}\]

Then just take the limit of it as n approaches infinity