Prove that $e^\frac{\pi}{2}=1+\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{5}{24}+\frac{1}{6}+\frac{85}{720}+\frac{520}{5040}+\frac{3145}{40320}+\cdots$

Question

I misused the Taylor series to represent $e^x$ as a combination of sin(x) functions

Assume: $e^x=a_0+a_1\sin(x)+a_2\sin^2(x)+a_3\sin^3(x)+\cdots$

The sequence coefficients are:$a_0=1;a_1=1;a_2=\frac{1}{2};a_3=\frac{2}{6};a_4=\frac{5}{24};a_5=\frac{20}{120};a_6=\frac{85}{720};a_7=\frac{520}{5040};a_8=\frac{3145}{40320}\cdots$

Therefore we have,

$e^\tfrac{\pi}{2}=1+\frac{1}{1}+\frac{1}{2}+\frac{2}{6}+\frac{5}{24}+\frac{20}{120}+\frac{85}{720}+\frac{520}{5040}+\frac{3145}{40320}+\cdots$

Additionally we have, $e^\tfrac{\pi}{6}=1+\frac{1}{2}+\frac{1}{2\times2^2}+\frac{2}{6\times2^3}+\frac{5}{24\times2^4}+\frac{20}{120\times2^5}+\frac{85}{720\times2^6}+\frac{518}{5040\times2^7}+\frac{3145}{40320\times2^8}+\cdots$

The summation values do seem to be nearly equal to the LHS after a few terms. However I am skeptical and would like to know if this misrepresentation of the Taylor series is valid.

I would appreciate any other proofs for the derived value of $e^\tfrac{\pi}{2}$ or $e^\tfrac{\pi}{6} $ to corroborate the summation.

Answer 1

If $F(z) = a_0 + a_1 z + a_2 z^2 + a_3 z^3 + \dots$, then your equation is $$ e^x = F(\sin x) . $$ So, solving, $$ F(z) = \exp(\arcsin z) . $$ near $z=0$.
Maple computes this Taylor series $$ \exp(\arcsin z) = 1+z+{\frac{1}{2}}{z}^{2}+{\frac{1}{3}}{z}^{3}+{\frac{5}{24}}{z}^{4}+{ \frac{1}{6}}{z}^{5}+{\frac{17}{144}}{z}^{6}+{\frac{13}{126}}{z}^{7}+{ \frac{629}{8064}}{z}^{8}+\dots $$ and this agrees with your values.

Answer 2

Just a remark :

Use $x>1$:

$$e^{\arctan\left(x\right)}<\int_{x}^{x+1}e^{\arctan\left(y\right)}dy<e^{\arctan\left(x+1\right)}$$

Then use Riemann sum.

Another attempt should be to use :

$$\prod_{k=0}^{\infty}\frac{\left(2k+2\right)^{2}}{\left(2k+1\right)\left(2k+3\right)}=\frac{\pi}{2}$$

Edit :

I have now a strategy let me share it :

1. show for $x,y\in[1,\infty)$ :

$$|xe^{\arctan(x)}-ye^{\arctan(y)}|\leq \exp(\pi/2)|x-y|$$

2)Show there exists a small $\varepsilon>0$ such that :

$$\int_{x}^{x+1}ye^{\arctan(y)}dy +\frac{1}{2}\exp(\pi/2)-\varepsilon\leq (x+1)e^{\arctan(x+1)}$$

3)Find a similar upper bound as in 2)

4)Make the difference in 2) and 3) and use Riemann's sum

5)Final step using @TymaGaidash finding ( OEIS A006228) show that if $AB=k$ and $AC=k$ it implies $B=C$

Answer 3

Starting from @GEdgar's answer, you could as many terms as you wish since $$F(z) = \exp(\arcsin( z))=\sum_{n=0}^\infty a_n\,z^n$$ with $$a_n=\frac{\left(n^2-4 n+5\right) }{(n-1) n}\,a_{n-2}\qquad \text{with}\qquad a_0=a_1=1$$

Computing the partial sums with $z=1$ $$S_p=\sum_{n=0}^\infty a_n$$ you will need a lot of terms since $\frac{a_{n+1}}{a_n}$ happens to be larger than $1$; so, for $z=1$, it will probably not converge. Converted to decimals, the results are

$$\left( \begin{array}{cc} p & S_p \\ 10 & 3.63555 \\ 20 & 3.96602 \\ 30 & 4.11724 \\ 40 & 4.20849 \\ 50 & 4.27117 \\ 60 & 4.31763 \\ 70 & 4.35384 \\ 80 & 4.38308 \\ 90 & 4.40734 \\ 100 & 4.42789 \\ 200 & 4.53951 \\ 300 & 4.58912 \\ 400 & 4.61872 \\ 500 & 4.63894 \\ 600 & 4.65387 \\ 700 & 4.66547 \\ 800 & 4.67483 \\ 900 & 4.68258 \\ 1000 & 4.68914 \\ 2000 & 4.72467 \\ 3000 & 4.74041 \\ \end{array} \right)$$

But, using $z=\frac 9 {10}$, it does converge like a charm

$$\left( \begin{array}{cc} p & S_p \\ 10 & 2.97617 \\ 20 & 3.04968 \\ 30 & 3.06103 \\ 40 & 3.06339 \\ 50 & 3.06395 \\ 60 & 3.06409 \\ 70 & 3.06413 \\ 80 & 3.06414 \\ 90 & 3.06415 \\ 100 & 3.06415 \\ \end{array} \right)$$