Show $ 4\sum_{k=0}^\infty \frac{(-1)^k (k+2)(k+1)}{(2k+5)!}x^{2k+2}=\frac{3}{x^3}\sin(x)-\frac{1}{x}\sin(x)-\frac{3}{x^2}\cos(x) $

Question

I'm trying to prove the following equality.$$ 4\sum_{k=0}^\infty \frac{(-1)^k (k+2)(k+1)}{(2k+5)!}x^{2k+2}=\frac{3}{x^3}\sin(x)-\frac{1}{x}\sin(x)-\frac{3}{x^2}\cos(x) $$

I thought it might be easier to start from the RHS, so what I have so far is:

$\frac{3}{x^3}\sin(x)-\frac{1}{x}\sin(x)-\frac{3}{x^2}\cos(x)$

=$\frac{3}{x^3}\sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!}x^{2k+1}-\frac{1}{x}\sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!}x^{2k+1}-\frac{3}{x^2}\sum_{k=0}^\infty \frac{(-1)^k}{(2k)!}x^{2k}$

$=3\sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!}x^{2k-2}-\sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!}x^{2k}-3\sum_{k=0}^\infty \frac{(-1)^k}{(2k)!}x^{2k-2}$

$=3\sum_{k=0}^\infty (\frac{(-1)^k}{(2k+1)!}-\frac{(-1)^k}{(2k)!})x^{2k-2}-\sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!}x^{2k}$

but I don't know where to go from here and would really appreciate if someone could show me how you get it in terms of sines and cosines?

Answer 1

You can easily calculate and simplify the left sum:

$$\sum_{k=0}^\infty (\frac{(-1)^k}{(2k+1)!}-\frac{(-1)^k}{(2k)!})x^{2k-1}$$

$$\sum_{k=0}^\infty (\frac{1}{(2k+1)!}-\frac{1}{(2k)!})(-1)^kx^{2k-1}$$

you can now calculate what is inside the first sum since 2k+1! = (2k+1)2k!

$$\sum_{k=0}^\infty (\frac{1}{(2k+1)!}-\frac{(2k+1)}{(2k+1)!})(-1)^kx^{2k-1}$$

$$\sum_{k=0}^\infty (\frac{-2k}{(2k+1)!})(-1)^kx^{2k-1}$$

Answer 2

The two expressions are not equal because $$ \dfrac{3}{x^2}\sin x-\dfrac{1}{x}\sin x-\dfrac{3}{x^2}\cos x=\frac{1}{2}-\dfrac{3}{x^2}+\dfrac{3}{x}-\dfrac{x}{2}+\dfrac{x^2}{6}+o(x) $$ but $$ \lim_{x\to0}\text{LHS}=0, $$

Answer 3

$$\frac{3}{x^3}\sin(x)-\frac{1}{x}\sin(x)-\frac{3}{x^2}\cos(x)$$

we know that

$$\sin (x)=\sum _{k=0}^{\infty } \frac{(-1)^k x^{2 k+1}}{(2 k+1)!}\text{ so we have }\frac{3 \sin (x)}{x^3}=3 \sum _{k=0}^{\infty } \frac{(-1)^k x^{2 k-2}}{(2 k+1)!}$$ and $$\sin (x)=\sum _{k=0}^{\infty } \frac{(-1)^{k+1} x^{2 k-1}}{(2 k-1)!}\text { which gives } -\frac{\sin (x)}{x}=-\sum _{k=0}^{\infty } \frac{(-1)^{k+1} x^{2 k-2}}{(2 k-1)!}$$ and finally $$\cos (x)=\sum _{k=0}^{\infty } \frac{(-1)^k x^{2 k}}{(2 k)!}\text{ from which we have } -\frac{3 \cos (x)}{x^2}=-3 \sum _{k=0}^{\infty } \frac{(-1)^k x^{2 k-2}}{(2 k)!}$$ Adding the series we get $$\frac{3 \sin (x)}{x^3}-\frac{\sin (x)}{x}-\frac{3 \cos (x)}{x^2}=\\=3 \sum _{k=0}^{\infty } \frac{(-1)^k x^{2 k-2}}{(2 k+1)!}-\sum _{k=0}^{\infty } \frac{(-1)^{k+1} x^{2 k-2}}{(2 k-1)!}-3\sum _{k=0}^{\infty } \frac{ (-1)^k x^{2 k-2}}{(2 k)!}$$ let's look at the general terms only $$3\frac{(-1)^k x^{2 k-2}}{(2 k+1)!}-\ \frac{(-1)^{k+1} x^{2 k-2}}{(2 k-1)!}-3\frac{ (-1)^k x^{2 k-2}}{(2 k)!}$$ to get the same denominator $$\frac{3 (-1)^k x^{2 k-2}}{(2 k+1)!}+\frac{2 (-1)^k k (2 k+1) x^{2 k-2}}{(2 k+1)!}-\frac{3 (-1)^k (2 k+1) x^{2 k-2}}{(2 k+1)!}$$ collect common factor $$\frac{(-1)^k x^{2 k-2}}{(2 k+1)!}\left( 3+2k(2k+1)-3(2k+1)\right)$$ expand the content of the parenthesis $$\frac{(-1)^k x^{2 k-2}}{(2 k+1)!}\left[ 4 (-1 + k) k\right]$$ Now to get the result equal to the LHS we substitute $k\to k+2$ $$4\frac{(-1)^{k+2} x^{2 (k+2)-2}}{(2 (k+2)+1)!} (-1 + k+2) (k+2)$$ and concluding $$4\frac{(-1)^{k} x^{2 k+2}}{(2k+5)!} (k+1) (k+2)$$

Hope this helps