Special sum to show by hand : $S<e$

Question

Problem :

Show that :

$$S=\sum_{n=1}^{104}\arcsin\left(\frac{2}{n^2+1}\right)<e$$

Without a computer (by hand).

This problem seems very difficult.

To show it I have used Jordan's inequality :

Let $0<x<1$ then we have :

$$\pi/2x>\arcsin(x)$$

But it's too far.Hopefully we can truncate the series .

We have the inequality :

Let $x>11$ then we have :

$$g(x)=\arcsin\left(2/(x^2+1)\right)<f(x)=2/x^2+2/(1-x^4)+1/32*1/x^4$$

Then with a computer we have :

$$\sum_{n=1}^{10}g(n)+\sum_{n=11}^{104}f(n)<e$$

Then we can use :

$\exists m>0$ large enought such that :

$$\sum_{i=1}^{m}1/i^2<\pi^2/6-1/m+0.5*1/m^2\tag{I}$$

Better approach :

Using $0<x<1$ :

$$\operatorname{arctanh}(x)>\arcsin(x)$$

And cutting the series it telescopes.

We can improves it with for $0<x<1$:

$$(\operatorname{arctanh }(x))/2+x/2>\arcsin(x)$$

Cutting and use $I$.

How to show by hand?

Answer 1

Remarks: Recently, I learned the method of bounding the series by using telescoping series. Perhaps we may find telescoping series better than (2) and (5).

Some thoughts:

It suffices to prove that $$\sum_{n=1}^\infty \arcsin\frac{2}{n^2 + 1} < \mathrm{e} + \sum_{n=105}^\infty \arcsin\frac{2}{n^2 + 1}. \tag{1}$$

Using $$\arcsin\frac{2}{n^2 + 1} > \frac{2}{n - \frac{12}{25}} - \frac{2}{n + 1 - \frac{12}{25}}, \quad \forall n \ge 105, \tag{2}$$ (easy to prove by taking derivative) we have $$ \sum_{n=105}^\infty \arcsin\frac{2}{n^2 + 1} > \sum_{n=105}^\infty \left(\frac{2}{n - \frac{12}{25}} - \frac{2}{n + 1 - \frac{12}{25}}\right) = \frac{2}{105 - \frac{12}{25}} = \frac{50}{2613}. \tag{3}$$

Thus, it suffices to prove that $$\sum_{n=1}^\infty \arcsin\frac{2}{n^2 + 1} < \mathrm{e} + \frac{50}{2613}. \tag{4}$$

We have, for all $n \ge 7$, \begin{align*} \arcsin\frac{2}{n^2 + 1} < g(n) &:= 2\left(\frac{1}{n} - \frac{1}{n + 1}\right) + \left(\frac{1}{n^2} - \frac{1}{(n + 1)^2}\right)\\ &\qquad - \frac13 \left(\frac{1}{n^3} - \frac{1}{(n + 1)^3}\right) - \left(\frac{1}{n^4} - \frac{1}{(n + 1)^4}\right)\\ &\qquad - \frac{1}{15}\left(\frac{1}{n^5} - \frac{1}{(n + 1)^5}\right) + \frac53\left(\frac{1}{n^6} - \frac{1}{(n + 1)^6}\right)\\ &\qquad + \frac{25}{21}\left(\frac{1}{n^7} - \frac{1}{(n + 1)^7}\right). \tag{5} \end{align*} (It can be proved by taking derivative. The proof is quite complicated by hand.)

Using (5), we have \begin{align*} &\sum_{n=1}^{\infty} \arcsin\frac{2}{n^2 + 1}\\ <{}&\sum_{n=1}^{6} \arcsin\frac{2}{n^2 + 1} + \sum_{n=7}^{\infty} g(n)\\ ={}& \frac{\pi}{2} + \arcsin\frac25 + \arcsin\frac15 + \arcsin\frac{2}{17} + \arcsin\frac{1}{13} + \arcsin\frac{2}{37}\\ &\quad + 2\left(\frac{1}{7} \right) + \left(\frac{1}{7^2}\right) - \frac13 \left(\frac{1}{7^3} \right) - \left(\frac{1}{7^4} \right)- \frac{1}{15}\left(\frac{1}{7^5}\right) + \frac53\left(\frac{1}{7^6} \right) + \frac{25}{21}\left(\frac{1}{7^7} \right)\\ <{}& \mathrm{e} + \frac{50}{2613}. \end{align*}

We are done.