Question about an inequality during proof that $e^2$ is irrational.

Question

I am reading a proof of the irrationality of $e^2$ and I am stuck on the following inequality:
Let $S := -a\underbrace{\left(\frac{1}{n+1} - \frac{1}{(n+1)(n+2)} + \frac{1}{(n+1)(n+2)(n+3)} \mp ...\right)}_{S^*}$ (just because of space issues) with $a \in \mathbb{Z}$, $n \in \mathbb{N}$. The proof I am reading states that $$ -\frac{a}{n} < S < -a\underbrace{\left(\frac{1}{n+1} - \frac{1}{(n+1)^2} - \frac{1}{(n+1)^3} - ...\right)}_{\tilde S} = -\frac{a}{n+1}\left(1-\frac{1}{n}\right) < 0. $$ Why is this true? My intuition tells me that $S^* < \frac{1}{n}$ since $1/n$ is already greater than the first term of $S^*$ and the terms afterwards all tends to 0 rather quickly so it never reaches $1/n$ but I am looking for a more rigorous explanation. The same goes for $S^*$ being apparently smaller than $\tilde S$. I see that the terms which get subtracted tends to zero more quickly than the terms of $S^*$ and so the inequality could be true as far as my intuition goes but not further. Thanks in advance for any help!

(The proof I am referring to is out of "Proofs from THE BOOK" by Martin Aigner and Günter M. Ziegler in case anyone is wondering.)

Answer 1

It took me kind of long to understand this proof but I think I finally understood and want to share what I've learned at this point. Also I think the way I stated my question was kind of misleading so I apologize for that.
First of all we assumed that $e^2$ is in fact rational with $e^2 = \frac{a}{b}$ where $a$ and $b$ are positive integers (since $e^2$ is also definitely positive). We can write this equation as $bn!e = an!e^{-1}$ (we also multiplied by $n!$ for some $n \in \mathbb{N})$ and insert the definition of $e$ and $e^{-1}$ via the exponential function (in the following for $e^{-1}$ and $n$ even): $$ e^{-1} = \sum^\infty_{k=0} \frac{(-1)^k}{k!} = \left( 1 - \frac{1}{1!} + \frac{1}{2!} \mp ... + \frac{1}{n!}\right) - \left(\frac{1}{(n+1)!} - \frac{1}{(n+2)!} + \frac{1}{(n+3)!} \mp ...\right). $$ Doing this the right hand side of the equation breaks up into two pieces one of which is $$ -an!\left(\frac{1}{(n+1)!} - \frac{1}{(n+2)!} + \frac{1}{(n+3)!} \mp ...\right) = -a\biggl(\underbrace{\frac{1}{n+1} - \frac{1}{(n+1)(n+2)} + \frac{1}{(n+1)(n+2)(n+3)} \mp ...}_{S^*}\biggl). $$ Now we want to approximate the value of $S^*$ and this is where the above inequality comes from. I think I should have made this a lot clearer in my question. As @ParamanandSingh pointed out $S^*$ passes the Leibniz-test for an alternating series. This is very useful because the value of any alternating series which passes the test can be approximated via $S^*_{2k-1} \leq S^* \leq S^*_{2k} \ \forall k \in \mathbb{N}_0$ where $S^*_k$ is the $k$-th partial sum of $S^*$ hence for $k=1$ we get $$ S^* \leq \frac{1}{n+1} - \frac{1}{(n+1)(n+2)} + \frac{1}{(n+1)(n+2)(n+3)} < \frac{1}{n+1} + \frac{1}{n+1} < \frac{2}{n} $$ and $$ S^* \geq \frac{1}{n+1} - \frac{1}{(n+1)(n+2)} = \frac{n+1}{(n+1)(n+2)} = \frac{1}{n+2}. $$ So in total we have $$ -\frac{2a}{n} \leq -aS^* \leq -\frac{a}{n+2} < 0 $$ which is enough to conclude that for large enough $n$ the value of $-aS^*$ will lie inside the open interval $(-1, 0)$.

Answer 2

For $a > 0$, if you divide through by $(-a)$, the inequality becomes $$\frac1{n+1} - \sum_{k=1}^\infty \frac1{(n+1)^k} < \frac1{n+1} - \sum_{k=1}^\infty \frac{(-1)^{k+1}n!}{(n+1+k)!} < \frac 1n = \frac1{n+1} - \frac{-1}{n^2 + n}$$ Or equivalently,

$$\frac{-1}{n^2 + n} < \sum_{k=1}^\infty \frac{(-1)^{k+1}n!}{(n+1+k)!} < \sum_{k=1}^\infty \frac1{(n+1)^k} = \frac1n$$

The middle summation is alternating and the absolute value of the terms decreases to $0$ as $k \to \infty$. Thus it must converge.

If we sum the terms in pairs, each pair consists of a positive term added to a negative term of smaller size, so every partial sum is $> 0$. So the full summation is $\ge 0$. which shows the left hand inequality.

For the right hand inequality, we can rewrite the middle summation as $$\frac 1{n+1} - \sum_{k=2}^\infty \frac{(-1)^kn!}{(n+1+k)!}$$ Once again this is an alternating series of decreasing terms with positive first term, meaning the summation itself is also $> 0$, and thus the full expression is $< \frac1{n+1} < \frac 1n$. Thus the right hand inequality also holds.